Construct Binary Tree from Preorder and Inorder Traversal Given preorder and inorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. 递归构建. 思路就是: preorder可以定位到根结点,inorder可以定位左右子树的取值范围. 1. 由…

LeetCode:Construct Binary Tree from Inorder and Postorder Traversal Given inorder and postorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree.                                            …

/* * 105. Construct Binary Tree from Inorder and preorder Traversal * 11.20 By Mingyang * 千万不要以为root一定在中间那个点,还是要找一下中间点的位置 * p.left = construct(preorder, preStart + 1, preStart + (k - inStart),inorder, inStart, k - 1); * p.right = construct(preorder,…

Construct Binary Tree from Inorder and Postorder Traversal Given inorder and postorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. Hide Tags Tree Array Depth-first Search   SOLUTION 1:…

Sept. 13, 2015 Spent more than a few hours to work on the leetcode problem, and my favorite blogs about this problems: 1. http://siddontang.gitbooks.io/leetcode-solution/content/tree/construct_binary_tree.html 2.http://blog.csdn.net/linhuanmars/artic…

Given inorder and postorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. 这道题要求从中序和后序遍历的结果来重建原二叉树,我们知道中序的遍历顺序是左-根-右,后序的顺序是左-右-根,对于这种树的重建一般都是采用递归来做,可参见我之前的一篇博客Convert Sorted Array to Bin…

Given preorder and inorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. Solution: /** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; *…

Given inorder and postorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. class Solution { public: TreeNode *buildTree(vector<int>& inorder, int in_left,int in_right, vector<int&…

原题地址:http://oj.leetcode.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/ 题意:根据二叉树的中序遍历和后序遍历恢复二叉树. 解题思路:看到树首先想到要用递归来解题.以这道题为例:如果一颗二叉树为{1,2,3,4,5,6,7},则中序遍历为{4,2,5,1,6,3,7},后序遍历为{4,5,2,6,7,3,1},我们可以反推回去.由于后序遍历的最后一个节点就是树的根.也就是roo…

Given inorder and postorder traversal of a tree, construct the binary tree. Note:  You may assume that duplicates do not exist in the tree. 利用中序和后序遍历构造二叉树,要注意到后序遍历的最后一个元素是二叉树的根节点,而中序遍历中,根节点前面为左子树节点后面为右子树的节点.例如二叉树:{1,2,3,4,5,6,#}的后序遍历为4->5->2->6-&…