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…

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-&…

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 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 * public class TreeNode { * int val; * TreeNode left; * TreeNode righ…

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…

Question Given inorder and postorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. Solution 参考:http://www.cnblogs.com/zhonghuasong/p/7096300.html Code /** * Definition for a binary tree…

map<int, int> mapIndex; void mapToIndex(int inorder[], int n) { ; i < n; i++) { mapIndex.insert(map<int, int>::value_type(inorder[i], i); } } Node* buildInorderPreorder(int in[], int pre[], int n, int offset) { assert(n >= ); ) return NU…

Given preorder and inorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. 这道题要求用先序和中序遍历来建立二叉树,跟之前那道Construct Binary Tree from Inorder and Postorder Traversal 由中序和后序遍历建立二叉树原理基本相同,针对这道题,由于先…

Construct Binary Tree from Inorder and Postorder Traversal OJ: https://oj.leetcode.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/ Given inorder and postorder traversal of a tree, construct the binary tree. Note: You may assu…

LeetCode 原题链接 Construct Binary Tree from Inorder and Postorder Traversal - LeetCode Construct Binary Tree from Preorder and Postorder Traversal - LeetCode 题目大意 给定一棵二叉树的中序遍历和后序遍历,求这棵二叉树的结构. 给定一棵二叉树的前序遍历和中序遍历,求这棵二叉树的结构. 样例 Input: inorder = [9, 3, 15, 2…

Construct Binary Tree from Inorder and Postorder Traversal 题解 原创文章,拒绝转载 题目来源:https://leetcode.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/description/ Description Given inorder and postorder traversal of a tree, construct…

1.  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. 代码: class Solution { public: TreeNode *buildTr…

[LeetCode]106. Construct Binary Tree from Inorder and Postorder Traversal 解题报告(Python) 标签: LeetCode 题目地址:https://leetcode.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/description/ 题目描述: Given inorder and postorder traversal…

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. 根据定义,后序遍历postorder的最后一个元素为根. 由于元素不重复,通过根可以讲中序遍历inorde…

Construct Binary Tree from Inorder and Postorder Traversal Total Accepted: 31041 Total Submissions: 115870     Given inorder and postorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree.…

Construct Binary Tree from Inorder and Postorder Traversal Given inorder and postorder traversal of a tree, construct the binary tree. 根据后序遍历和中序遍历构建一棵二叉树 /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * Tree…

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. SOLUTION 1: 1. Find the root node from the preorder.(it…

一天一道LeetCode 本系列文章已全部上传至我的github,地址:ZeeCoder's Github 欢迎大家关注我的新浪微博,我的新浪微博 欢迎转载,转载请注明出处 (一)题目 来源:https://leetcode.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/ Given inorder and postorder traversal of a tree, construct the b…

Given inorder and postorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. For example, given inorder = [9,3,15,20,7] postorder = [9,15,7,20,3] Return the following binary tree: 3 / \ 9 2…

Given inorder and postorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. For example, given inorder = [9,3,15,20,7] postorder = [9,15,7,20,3] Return the following binary tree: 3 / \ 9 2…

Given inorder and postorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. 这道题之前算法课上好像遇到过,思路也很简单的. 思路:后序序列的最后一个元素就是树根,然后在中序序列中找到这个元素(由于题目保证没有相同的元素,因此可以唯一找到),中序序列中这个元素的左边就是左子树的中序,右边就是右子树的…

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

Given preorder and inorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. 此题目有两种解决思路: 1)递归解决(比较好想)按照手动模拟的思路即可 2)非递归解决,用stack模拟递归 class Solution { public: TreeNode *buildTree(vector<int>&…

Given preorder and inorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. 难度:95,参考了网上的思路.这道题是树中比较有难度的题目,需要根据先序遍历和中序遍历来构造出树来.这道题看似毫无头绪,其实梳理一下还是有章可循的.下面我们就用一个例子来解释如何构造出树.假设树的先序遍历是12453687,…