105. Construct Binary Tree from Preorder and Inorder Traversal
作者:互联网
Given preorder and inorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
见剑指offer重建二叉树
class Solution { public TreeNode buildTree(int[] preorder, int[] inorder) { return helper(0, 0, inorder.length - 1, preorder, inorder); } public TreeNode helper(int preStart, int inStart, int inEnd, int[] preorder, int[] inorder) { if (preStart > preorder.length - 1 || inStart > inEnd) return null; TreeNode root = new TreeNode(preorder[preStart]); int inIndex = 0; for (int i = inStart; i <= inEnd; i++) { if (inorder[i] == root.val) inIndex = i; } root.left = helper(preStart + 1, inStart, inIndex - 1, preorder, inorder); root.right = helper(preStart + inIndex - inStart + 1, inIndex + 1, inEnd, preorder, inorder); return root; } }标签:Binary,Preorder,preorder,int,Tree,inIndex,inStart,preStart,inorder 来源: https://www.cnblogs.com/MarkLeeBYR/p/10536839.html