package LeetCode_105 /** * 105. Construct Binary Tree from Preorder and Inorder Traversal * https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/description/ * * Given preorder and inorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. For example, given preorder = [3,9,20,15,7] inorder = [9,3,15,20,7] Return the following binary tree: 3 / 9 20 / 15 7 * */ class Solution { class TreeNode(var `val`: Int = 0) { var left: TreeNode? = null var right: TreeNode? = null } /* Time complexity:O(n^2), Space complexity:O(n) * preorder: root->left->right * inorder: left->root->right * so the first element of perorder array is the root of the tree, * then divide the inorder array based in the the first element of preorder array that two part will be left and right * */ var rootIndex = 0 fun buildTree(preorder: IntArray, inorder: IntArray): TreeNode? { rootIndex = 0 return buildTree(preorder,0,inorder.size-1,inorder) } private fun buildTree(preorder: IntArray, start:Int, end:Int, inorder: IntArray):TreeNode?{ if (start>end){ return null } if (start==end){ return TreeNode(preorder[rootIndex++]) } val root = TreeNode(preorder[rootIndex++]) for (i in start..end){ if (inorder[i]==root.`val`){ //if is preorder array, we build left node first //if is postorder array, we build right node first root.left = buildTree(preorder, start, i-1, inorder) root.right = buildTree(preorder, i+1, end, inorder) return root } } return root } }