title

L48. Construct a BST from a preorder traversal | 3 Methods

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{'title': 'L48. Construct a BST from a preorder traversal | 3 Methods', 'heatmap': [{'end': 823.843, 'start': 800.655, 'weight': 0.725}, {'end': 933.978, 'start': 863.439, 'weight': 0.755}, {'end': 973.356, 'start': 951.309, 'weight': 0.73}], 'summary': 'Covers three chapters on constructing binary search trees, including methods for creating bst from pre-order and in-order traversals, discussing time and space complexities, and highlighting efficient algorithms with o(n) time complexity.', 'chapters': [{'end': 178.259, 'segs': [{'end': 105.922, 'src': 'embed', 'start': 39.317, 'weight': 0, 'content': [{'end': 48.441, 'text': 'if i take this example and you have to draw the corresponding binary search tree for this given pre-order and once you have created this uh corresponding binary search tree,', 'start': 39.317, 'duration': 9.124}, {'end': 52.522, 'text': 'you have to just return the root of that binary search tree.', 'start': 48.441, 'duration': 4.081}, {'end': 54.984, 'text': 'yes, that is what you have to do.', 'start': 52.522, 'duration': 2.462}, {'end': 56.526, 'text': 'so how will you solve this problem?', 'start': 54.984, 'duration': 1.542}, {'end': 58.668, 'text': "what's the brute way to solve this problem?", 'start': 56.526, 'duration': 2.142}, {'end': 64.235, 'text': 'the brute way to solve this problem is to manually, like naively, create the binary tree.', 'start': 58.668, 'duration': 5.567}, {'end': 66.497, 'text': 'now, when i say naively, what does that signify?', 'start': 64.235, 'duration': 2.262}, {'end': 67.599, 'text': "let's check that out.", 'start': 66.497, 'duration': 1.102}, {'end': 69.601, 'text': 'so initially you have eight.', 'start': 67.599, 'duration': 2.002}, {'end': 71.403, 'text': 'now you know what is pre-order?', 'start': 69.601, 'duration': 1.802}, {'end': 76.047, 'text': 'pre-order is basically root node left, right.', 'start': 71.403, 'duration': 4.644}, {'end': 79.148, 'text': 'that is what we call as pre-order traversal.', 'start': 76.047, 'duration': 3.101}, {'end': 83.429, 'text': 'so we definitely know that this is the root of the binary search tree right.', 'start': 79.148, 'duration': 4.281}, {'end': 86.29, 'text': 'so we can say that this is the root of the binary search tree.', 'start': 83.429, 'duration': 2.861}, {'end': 87.63, 'text': 'next we come across to 5.', 'start': 86.29, 'duration': 1.34}, {'end': 94.552, 'text': 'now, whenever we come across to 5, we know everything to the left of 8 should be smaller than 8.', 'start': 87.63, 'duration': 6.922}, {'end': 96.934, 'text': 'so five can be inserted over here.', 'start': 94.552, 'duration': 2.382}, {'end': 100.557, 'text': 'so five is also inserted next time we come across to one.', 'start': 96.934, 'duration': 3.623}, {'end': 105.922, 'text': 'now, when we come across to one, we know everything to the left of eight is smaller than eight.', 'start': 100.557, 'duration': 5.365}], 'summary': 'Given a pre-order sequence, construct a binary search tree and return its root.', 'duration': 66.605, 'max_score': 39.317, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/UmJT3j26t1I/pics/UmJT3j26t1I39317.jpg'}, {'end': 187.383, 'src': 'embed', 'start': 156.682, 'weight': 1, 'content': [{'end': 158.002, 'text': 'where do we need to insert it?', 'start': 156.682, 'duration': 1.32}, {'end': 159.803, 'text': "you're always determining that right.", 'start': 158.002, 'duration': 1.801}, {'end': 164.644, 'text': "so i can say you will be taking a big of n time if it's a skew tree like.", 'start': 159.803, 'duration': 4.841}, {'end': 172.597, 'text': "if you're given a tree like this, then you'll end up taking an entire lot of time to insert every particular note.", 'start': 164.644, 'duration': 7.953}, {'end': 178.259, 'text': "So the worst case, that can be bico of n into n, and there is no external space that you're using,", 'start': 172.837, 'duration': 5.422}, {'end': 182.981, 'text': "apart from the one that you're using to create this tree, which is needed.", 'start': 178.259, 'duration': 4.722}, {'end': 184.722, 'text': 'So we do not consider that.', 'start': 183.301, 'duration': 1.421}, {'end': 187.383, 'text': 'So this will be the absolute root method.', 'start': 185.022, 'duration': 2.361}], 'summary': 'Inserting into a skew tree can take o(n^2) time with no external space used.', 'duration': 30.701, 'max_score': 156.682, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/UmJT3j26t1I/pics/UmJT3j26t1I156682.jpg'}], 'start': 0.149, 'title': 'Constructing binary search tree', 'summary': 'Discusses constructing a binary search tree from pre-order traversal, demonstrating the brute force method and highlighting the time complexity of the approach.', 'chapters': [{'end': 178.259, 'start': 0.149, 'title': 'Constructing binary search tree from pre-order traversal', 'summary': 'Discusses how to construct a binary search tree from a given pre-order traversal, demonstrating the brute force method and highlighting the time complexity of the approach.', 'duration': 178.11, 'highlights': ["Binary search tree follows the rule that everything to the left of a node is smaller than the node's value, and everything to the right is greater, creating a unique characteristic. Explains the defining characteristic of a binary search tree.", 'The problem involves creating a binary search tree from a given pre-order traversal and returning the root of the tree. Describes the specific task of creating a binary search tree from pre-order traversal.', 'The brute force method involves naively creating the binary search tree by determining the placement of each node based on the pre-order traversal. Explains the brute force method of creating the binary search tree from pre-order traversal.', 'The time complexity for the brute force method is O(n^2) in the worst case scenario, where n represents the number of nodes in the tree. Discusses the time complexity of the brute force method for creating the binary search tree.', 'The worst case time complexity can be O(n^2) if the tree is a skew tree, resulting in significant time consumption for inserting each node. Highlights the worst case time complexity for a skew tree in the brute force method.']}], 'duration': 178.11, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/UmJT3j26t1I/pics/UmJT3j26t1I149.jpg', 'highlights': ['The brute force method involves naively creating the binary search tree by determining the placement of each node based on the pre-order traversal.', 'The time complexity for the brute force method is O(n^2) in the worst case scenario, where n represents the number of nodes in the tree.', 'The worst case time complexity can be O(n^2) if the tree is a skew tree, resulting in significant time consumption for inserting each node.', "Binary search tree follows the rule that everything to the left of a node is smaller than the node's value, and everything to the right is greater, creating a unique characteristic.", 'The problem involves creating a binary search tree from a given pre-order traversal and returning the root of the tree.']}, {'end': 309.376, 'segs': [{'end': 229.598, 'src': 'embed', 'start': 204.216, 'weight': 1, 'content': [{'end': 214.028, 'text': 'so can i say if the given pre-order traversal consists of all the nodes of bst, if it consists of all the nodes of bst, so if i sort them,', 'start': 204.216, 'duration': 9.812}, {'end': 214.909, 'text': 'will i get the in order?', 'start': 214.028, 'duration': 0.881}, {'end': 218.672, 'text': "think logically, yes, so why don't I sort them?", 'start': 215.49, 'duration': 3.182}, {'end': 220.333, 'text': 'so I will sort them.', 'start': 218.672, 'duration': 1.661}, {'end': 226.836, 'text': 'then the in order will be 1, I think 1, 5, 7, 8 and 12.', 'start': 220.333, 'duration': 6.503}, {'end': 229.598, 'text': 'that is what the in order will be.', 'start': 226.836, 'duration': 2.762}], 'summary': 'Sorting pre-order traversal of bst yields in-order sequence: 1, 5, 7, 8, 12.', 'duration': 25.382, 'max_score': 204.216, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/UmJT3j26t1I/pics/UmJT3j26t1I204216.jpg'}, {'end': 280.13, 'src': 'embed', 'start': 251.75, 'weight': 0, 'content': [{'end': 258.113, 'text': "so in case you want to know how to create a binary tree, if you're given pre-order as well as in order,", 'start': 251.75, 'duration': 6.363}, {'end': 262.096, 'text': 'you can check my video out in this particular playlist only.', 'start': 258.113, 'duration': 3.983}, {'end': 267.043, 'text': 'okay, so can say that the time complexity will be okay.', 'start': 262.096, 'duration': 4.947}, {'end': 271.045, 'text': 'i will have to sort the pre-order in order to get the in order,', 'start': 267.043, 'duration': 4.002}, {'end': 280.13, 'text': 'so that will take n login and then i will take another big of n in order to create the bst from the pre-order as well as the in order and the space.', 'start': 271.045, 'duration': 9.085}], 'summary': 'Learn to create a binary tree with pre-order and in-order data, with a time complexity of o(n log n) and space complexity of o(n).', 'duration': 28.38, 'max_score': 251.75, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/UmJT3j26t1I/pics/UmJT3j26t1I251750.jpg'}], 'start': 178.259, 'title': 'Creating binary search tree from pre-order and in-order', 'summary': 'Discusses creating a binary search tree (bst) from pre-order and in-order traversals, including methods involving sorting and the efficient method with a time complexity of o(n).', 'chapters': [{'end': 309.376, 'start': 178.259, 'title': 'Creating binary search tree from pre-order and in-order', 'summary': 'Discusses creating a binary search tree (bst) from pre-order and in-order traversals, including the method involving sorting and the efficient method with a time complexity of o(n).', 'duration': 131.117, 'highlights': ['The in order traversal of a Binary Search Tree (BST) will always be sorted, which can be utilized to create a unique BST from given pre-order and in-order traversals.', 'Sorting the pre-order traversal to obtain the in-order traversal will take O(n log n) time complexity and an additional O(n) space complexity.', 'An efficient method to create a BST from pre-order and in-order traversals can be achieved in O(n) time complexity.']}], 'duration': 131.117, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/UmJT3j26t1I/pics/UmJT3j26t1I178259.jpg', 'highlights': ['An efficient method to create a BST from pre-order and in-order traversals can be achieved in O(n) time complexity.', 'The in order traversal of a Binary Search Tree (BST) will always be sorted, which can be utilized to create a unique BST from given pre-order and in-order traversals.', 'Sorting the pre-order traversal to obtain the in-order traversal will take O(n log n) time complexity and an additional O(n) space complexity.']}, {'end': 823.843, 'segs': [{'end': 338.672, 'src': 'embed', 'start': 309.376, 'weight': 0, 'content': [{'end': 318.673, 'text': 'if you have seen my previous video, that is check if a bt, a binary search tree I am assuming you have seen that okay.', 'start': 309.376, 'duration': 9.297}, {'end': 321.295, 'text': "so if I've seen that now, you know one thing.", 'start': 318.673, 'duration': 2.622}, {'end': 324.179, 'text': "you know one thing if there's a node, there will.", 'start': 321.295, 'duration': 2.884}, {'end': 326.161, 'text': 'it will always belong to a range.', 'start': 324.179, 'duration': 1.982}, {'end': 338.672, 'text': 'that as the starting node will always belong to minus int mean like that is int mean to int max, and the left one will always be lesser than this guy.', 'start': 326.161, 'duration': 12.511}], 'summary': 'Explains how nodes in a binary search tree belong to a specific range.', 'duration': 29.296, 'max_score': 309.376, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/UmJT3j26t1I/pics/UmJT3j26t1I309376.jpg'}, {'end': 380.58, 'src': 'embed', 'start': 352.202, 'weight': 4, 'content': [{'end': 358.606, 'text': 'so this is the logic that we will be using in order to create the binary source tree right.', 'start': 352.202, 'duration': 6.404}, {'end': 361.068, 'text': "so let's see, how can we use that now?", 'start': 358.606, 'duration': 2.462}, {'end': 363.249, 'text': 'do we require to carry both the ranges?', 'start': 361.068, 'duration': 2.181}, {'end': 364.97, 'text': 'assume this to be l and r?', 'start': 363.249, 'duration': 1.721}, {'end': 369.453, 'text': 'do we require, do we need both the ranges like the lower bound and the upper bound?', 'start': 364.97, 'duration': 4.483}, {'end': 371.255, 'text': 'the answer to that will be no.', 'start': 369.453, 'duration': 1.802}, {'end': 374.517, 'text': 'if we just carry the upper bound, then I think it will work.', 'start': 371.255, 'duration': 3.262}, {'end': 380.58, 'text': "Why the lower bound is not required, I'll tell you that while I give you the algorithm.", 'start': 374.957, 'duration': 5.623}], 'summary': 'Creating binary source tree logic discussed. only upper bound needed, lower bound not required.', 'duration': 28.378, 'max_score': 352.202, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/UmJT3j26t1I/pics/UmJT3j26t1I352202.jpg'}, {'end': 730.624, 'src': 'embed', 'start': 704.212, 'weight': 3, 'content': [{'end': 709.154, 'text': 'the upper bound is 10 and then 12 and then 12 is not possible.', 'start': 704.212, 'duration': 4.942}, {'end': 711.155, 'text': 'so you will be like not possible.', 'start': 709.154, 'duration': 2.001}, {'end': 712.156, 'text': "let's go back now.", 'start': 711.155, 'duration': 1.001}, {'end': 715.477, 'text': 'you go right and you say the upper bound is 8 max.', 'start': 712.156, 'duration': 3.321}, {'end': 716.558, 'text': 'so 12 is possible.', 'start': 715.477, 'duration': 1.081}, {'end': 721.64, 'text': 'here you place it and then you go back because there are no remaining elements.', 'start': 716.558, 'duration': 5.082}, {'end': 726.382, 'text': 'so ultimately you end up forming this entire BST.', 'start': 721.64, 'duration': 4.742}, {'end': 730.624, 'text': 'so you just have to carry this upper bound whenever you call left.', 'start': 726.382, 'duration': 4.242}], 'summary': 'Using upper bounds, form bst with 8 as max, resulting in the formed entire bst.', 'duration': 26.412, 'max_score': 704.212, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/UmJT3j26t1I/pics/UmJT3j26t1I704212.jpg'}, {'end': 826.845, 'src': 'heatmap', 'start': 800.655, 'weight': 1, 'content': [{'end': 806.819, 'text': 'So 5 is like, from up once, goes to the left, comes back, right, comes back.', 'start': 800.655, 'duration': 6.164}, {'end': 809.521, 'text': "Three times you're visiting a single node, at the worst case.", 'start': 807.119, 'duration': 2.402}, {'end': 818.898, 'text': "If you're visiting three times, can I say it's a 3N? which is as good as we go of n because you won't be visiting 3n always.", 'start': 810.141, 'duration': 8.757}, {'end': 820.88, 'text': "So that's as good as we go of n.", 'start': 818.938, 'duration': 1.942}, {'end': 823.843, 'text': 'And the space complexity, yeah, recursion is using a stack space.', 'start': 820.88, 'duration': 2.963}, {'end': 826.845, 'text': 'But yeah, not any external containers.', 'start': 824.343, 'duration': 2.502}], 'summary': 'Algorithm visits single node 3 times at worst case, leading to a time complexity of o(3n) and space complexity using stack space.', 'duration': 26.19, 'max_score': 800.655, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/UmJT3j26t1I/pics/UmJT3j26t1I800655.jpg'}], 'start': 309.376, 'title': 'Binary search trees', 'summary': 'Discusses the concept of node ranges in a binary search tree and explains the logic for creating a binary search tree, including time and space complexities.', 'chapters': [{'end': 352.202, 'start': 309.376, 'title': 'Binary search tree node range', 'summary': "Discusses the concept of node ranges in a binary search tree, where a node's value is always within a certain range, with the left node being lesser and the right node being greater, from the starting node to int max and min values.", 'duration': 42.826, 'highlights': ['Nodes in a binary search tree always belong to a specific range, with the starting node belonging to the range from int min to int max, the left node belonging to the range from int min to the upper node value, and the right node belonging to the range from the node value to int max.', 'Understanding the concept that nodes in a binary search tree belong to a specific range is crucial for traversal and manipulation of the tree.']}, {'end': 823.843, 'start': 352.202, 'title': 'Creating binary search tree logic', 'summary': 'Explains the logic for creating a binary search tree, illustrating the process with an example and discussing the reasons for not requiring a lower bound. the time complexity is approximately o(n) and the space complexity is determined by the recursion using a stack space.', 'duration': 471.641, 'highlights': ['The upper bound is used to determine the placement of elements in the binary search tree, and the process involves traversing through the elements using the upper bound as a reference for comparison.', 'The logic for not requiring a lower bound is explained through the example, where the inability to place an element automatically checks for the lower bound, simplifying the process of building the tree.', 'The time complexity is analyzed to be approximately O(n), with the explanation that each node is visited at most three times in the worst case scenario, simplifying the assessment of the overall traversal complexity.', 'The space complexity is determined by the recursion using a stack space, with the discussion emphasizing that the worst case scenario of visiting a node three times is still manageable and equivalent to O(n).']}], 'duration': 514.467, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/UmJT3j26t1I/pics/UmJT3j26t1I309376.jpg', 'highlights': ['Understanding the concept that nodes in a binary search tree belong to a specific range is crucial for traversal and manipulation of the tree.', 'The time complexity is analyzed to be approximately O(n), with the explanation that each node is visited at most three times in the worst case scenario, simplifying the assessment of the overall traversal complexity.', 'The space complexity is determined by the recursion using a stack space, with the discussion emphasizing that the worst case scenario of visiting a node three times is still manageable and equivalent to O(n).', 'The upper bound is used to determine the placement of elements in the binary search tree, and the process involves traversing through the elements using the upper bound as a reference for comparison.', 'The logic for not requiring a lower bound is explained through the example, where the inability to place an element automatically checks for the lower bound, simplifying the process of building the tree.']}, {'end': 991.493, 'segs': [{'end': 933.978, 'src': 'heatmap', 'start': 841.487, 'weight': 0, 'content': [{'end': 848.971, 'text': 'So as you can see that we will be having a vector in C++ and then Java, it will be integer.', 'start': 841.487, 'duration': 7.484}, {'end': 851.673, 'text': 'So initially we have kept the pointer.', 'start': 849.291, 'duration': 2.382}, {'end': 853.093, 'text': 'What is the pointer??', 'start': 852.473, 'duration': 0.62}, {'end': 854.974, 'text': 'Now, if you remember, we had a pre-order right?', 'start': 853.414, 'duration': 1.56}, {'end': 860.938, 'text': 'So initially we had a pointer like the first element that we compared with bound, then we had the next element, then we had the next.', 'start': 855.315, 'duration': 5.623}, {'end': 862.879, 'text': 'So we need a pointer, right, to move across.', 'start': 861.278, 'duration': 1.601}, {'end': 870.024, 'text': 'So, we have a pointer as i and we say build, we have passed on the vector, the i as well as the upper bound.', 'start': 863.439, 'duration': 6.585}, {'end': 873.606, 'text': 'Initially, when you start off at the root, the upper bound is int max.', 'start': 870.364, 'duration': 3.242}, {'end': 879.03, 'text': 'So whenever we reach and we say that okay, hey listen, if we ran out of elements like,', 'start': 874.086, 'duration': 4.944}, {'end': 882.232, 'text': 'there are no more elements that has to be inserted into the BST.', 'start': 879.03, 'duration': 3.202}, {'end': 889.097, 'text': "So, at that case, that's over or if it is out of bound, we will say, okay, this is not possible over here.", 'start': 882.892, 'duration': 6.205}, {'end': 889.997, 'text': 'You can go back.', 'start': 889.417, 'duration': 0.58}, {'end': 892.019, 'text': 'So, return a null in both the cases.', 'start': 890.397, 'duration': 1.622}, {'end': 897.863, 'text': 'and normally what you can say is, if this is not the case, that means it is okay.', 'start': 893.061, 'duration': 4.802}, {'end': 905.927, 'text': "so you'll take that guy at the ith index and you will say that hey listen, this will be a part of my current node.", 'start': 897.863, 'duration': 8.064}, {'end': 913.651, 'text': "like i'll form a root with that and once i form that, i know that roots left will go with root.val as the upper bound.", 'start': 905.927, 'duration': 7.724}, {'end': 916.772, 'text': 'but if it comes back, if it comes back,', 'start': 914.591, 'duration': 2.181}, {'end': 921.993, 'text': 'then it has to go to the right and for the right the bound will remain the same only whenever it goes to the left.', 'start': 916.772, 'duration': 5.221}, {'end': 929.535, 'text': 'the bound has to be, has to be self-adjusted, because whenever it goes to the right if you remember the example the moment it went to the right,', 'start': 921.993, 'duration': 7.542}, {'end': 933.978, 'text': 'intermax remained as intermax Correct?, Whereas 5 it remained as 5..', 'start': 929.535, 'duration': 4.443}], 'summary': 'Algorithm explanation of building a binary search tree using a vector in c++ and java, with pointer and upper bound considerations.', 'duration': 40.745, 'max_score': 841.487, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/UmJT3j26t1I/pics/UmJT3j26t1I841487.jpg'}, {'end': 982.583, 'src': 'heatmap', 'start': 951.309, 'weight': 0.73, 'content': [{'end': 955.17, 'text': 'now, in java, you cannot pass a reference to a variable.', 'start': 951.309, 'duration': 3.861}, {'end': 959.451, 'text': "so what i've done is i've created a array of one size yes,", 'start': 955.17, 'duration': 4.281}, {'end': 967.174, 'text': "an array of size one and since array can be passed by reference and i've used that array in order, like at its first element as it like,", 'start': 959.451, 'duration': 7.723}, {'end': 973.356, 'text': 'the at its first index, or the zeroth index as a variable, because variables cannot be passed.', 'start': 967.174, 'duration': 6.182}, {'end': 976.737, 'text': 'so, guys, i hope you have understood all the three explanations as well as the code.', 'start': 973.356, 'duration': 3.381}, {'end': 982.583, 'text': "just in case you did, please, please, please make sure you like this video and if you're new to this channel, please do consider subscribing,", 'start': 976.997, 'duration': 5.586}], 'summary': 'In java, a reference to a variable cannot be passed, so an array of size one is created and used to pass by reference instead.', 'duration': 31.274, 'max_score': 951.309, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/UmJT3j26t1I/pics/UmJT3j26t1I951309.jpg'}], 'start': 824.343, 'title': 'Bst node insertion algorithm', 'summary': 'Discusses the time and space complexity of bst node insertion algorithm, with insights into c++ and java implementation, pointers, and array references.', 'chapters': [{'end': 991.493, 'start': 824.343, 'title': 'Bst node insertion algorithm', 'summary': 'Discusses the time and space complexity of the bst node insertion algorithm, providing insights into the c++ and java implementation, including the usage of pointers and array references.', 'duration': 167.15, 'highlights': ['The chapter explains the time and space complexity of the BST node insertion algorithm, concluding that the presented approach is the most efficient.', 'The C++ and Java implementations of the algorithm are compared, highlighting the usage of pointers in C++ and array references in Java for managing the traversal and insertion process.', 'The detailed explanation includes the utilization of pointers to traverse the BST nodes in C++, while in Java, an array of size one is used to mimic the behavior of passing variables by reference.']}], 'duration': 167.15, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/UmJT3j26t1I/pics/UmJT3j26t1I824343.jpg', 'highlights': ['The chapter explains the time and space complexity of the BST node insertion algorithm, concluding that the presented approach is the most efficient.', 'The C++ and Java implementations of the algorithm are compared, highlighting the usage of pointers in C++ and array references in Java for managing the traversal and insertion process.', 'The detailed explanation includes the utilization of pointers to traverse the BST nodes in C++, while in Java, an array of size one is used to mimic the behavior of passing variables by reference.']}], 'highlights': ['An efficient method to create a BST from pre-order and in-order traversals can be achieved in O(n) time complexity.', 'The in order traversal of a Binary Search Tree (BST) will always be sorted, which can be utilized to create a unique BST from given pre-order and in-order traversals.', 'The time complexity is analyzed to be approximately O(n), with the explanation that each node is visited at most three times in the worst case scenario, simplifying the assessment of the overall traversal complexity.', 'The space complexity is determined by the recursion using a stack space, with the discussion emphasizing that the worst case scenario of visiting a node three times is still manageable and equivalent to O(n).', 'The upper bound is used to determine the placement of elements in the binary search tree, and the process involves traversing through the elements using the upper bound as a reference for comparison.', 'The logic for not requiring a lower bound is explained through the example, where the inability to place an element automatically checks for the lower bound, simplifying the process of building the tree.', 'The chapter explains the time and space complexity of the BST node insertion algorithm, concluding that the presented approach is the most efficient.', 'The C++ and Java implementations of the algorithm are compared, highlighting the usage of pointers in C++ and array references in Java for managing the traversal and insertion process.']}