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105. Construct Binary Tree from Preorder and Inorder Traversal
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| # https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/ | |
| # Given two integer arrays preorder and inorder where preorder is the preorder traversal of a binary tree and inorder is the inorder traversal of the same tree, construct and return the binary tree. | |
| # Example 1: | |
| # Input: preorder = [3,9,20,15,7], inorder = [9,3,15,20,7] | |
| # Output: [3,9,20,null,null,15,7] | |
| # Example 2: | |
| # Input: preorder = [-1], inorder = [-1] | |
| # Output: [-1] | |
| # Constraints: | |
| # 1 <= preorder.length <= 3000 | |
| # inorder.length == preorder.length | |
| # -3000 <= preorder[i], inorder[i] <= 3000 | |
| # preorder and inorder consist of unique values. | |
| # Each value of inorder also appears in preorder. | |
| # preorder is guaranteed to be the preorder traversal of the tree. | |
| # inorder is guaranteed to be the inorder traversal of the tree. | |
| # Definition for a binary tree node. | |
| # class TreeNode | |
| # attr_accessor :val, :left, :right | |
| # def initialize(val = 0, left = nil, right = nil) | |
| # @val = val | |
| # @left = left | |
| # @right = right | |
| # end | |
| # end | |
| # @param {Integer[]} preorder | |
| # @param {Integer[]} inorder | |
| # @return {TreeNode} | |
| def build_tree(preorder, inorder) | |
| subtree(preorder, inorder)[:node] | |
| end | |
| # general strategy | |
| # https://www.geeksforgeeks.org/construct-tree-from-given-inorder-and-preorder-traversal/ | |
| # https://www.geeksforgeeks.org/tree-traversals-inorder-preorder-and-postorder/ | |
| # given the preoder list, look for the first preorder item in the inorder list | |
| # use it as a node to split, build subtrees from the left and right lists | |
| # and use those as the left & right children of the node | |
| # return the node along with rest of preorder list for items that still need to be found | |
| def subtree(preorder, inorder) | |
| return {node: nil, rest: preorder} unless preorder | |
| # value check? | |
| return {node: TreeNode.new(preorder.first), rest: preorder[1..-1]} if inorder.size == 1 | |
| # another base case | |
| if inorder.size == 2 | |
| preorder2 = preorder.take(2) | |
| if preorder2 == inorder | |
| return { | |
| node: TreeNode.new(inorder.first, nil, TreeNode.new(inorder.last)), | |
| rest: preorder.drop(2) | |
| } | |
| else | |
| return { | |
| node: TreeNode.new(inorder.last, TreeNode.new(inorder.first), nil), | |
| rest: preorder.drop(2) | |
| } | |
| end | |
| end | |
| val, *rest = preorder | |
| node = TreeNode.new(val) | |
| result = {node:, rest:} | |
| # O(n^2) to O(n) improvement (for this part at least) | |
| # can use an item => index lookup hash of the inorder list | |
| # but need to update with offset of the reduced inorder list | |
| i = inorder.index(val) | |
| # look on the list on the left & right side of the split at node | |
| # except when the node is at the left/right end | |
| if preorder.first == inorder.first | |
| left, right = [], inorder.drop(1) | |
| elsif preorder.first == inorder.last | |
| left, right = inorder[0..-2], [] | |
| else | |
| left, right = inorder[0..i-1], inorder[i+1..-1] | |
| end | |
| # if there's no left or right subtrees, the left/right fields will be nil by default | |
| unless left.empty? | |
| result = subtree(rest, left) | |
| node.left = result[:node] | |
| end | |
| unless right.empty? | |
| result = subtree(result[:rest], right) | |
| node.right = result[:node] | |
| end | |
| {node:, rest: result[:rest]} | |
| end |
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