/* * 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,…

代码实现:给定一个中序遍历和后序遍历怎么构造出这颗树!(假定树中没有重复的数字) 因为没有规定是左小右大的树,所以我们随意画一颗数,来进行判断应该是满足题意的. 3 / \ 2 4 /\ / \1 6 5 7 中序遍历:. 后序遍历:. 我们知道后序遍历的最后一个肯定就是根了.然后在前序遍历中找到这个根,左边的就是左子树(记作sub),右边的就是右子树(记作sub).在后序遍历中,前面的几个对应左子树的后序遍历(记作sub),接下去的几个对应右子树的后序遍历(记作sub),注意,右子树的后序遍历…

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: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.                                            …

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 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…

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…

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 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:…

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 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…

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.…

[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…

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…

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. 提示: 题目要求通过一颗二叉树的中序遍历及后续遍历的结果,将这颗二叉树构建出来,另外还有一个已知条件,所有节点的值都是不同的. 首先需要了解一下二叉树不同遍历方式的定义: 前序遍历:首先访问根结点,然后遍历左子树,最…

总结: 1. 第 36 行代码, 最好是按照 len 来遍历, 而不是下标 代码: 前序中序 #include <iostream> #include <vector> using namespace std; struct TreeNode { int val; TreeNode *left; TreeNode *right; TreeNode(int x) : val(x), left(NULL), right(NULL) {} }; class Solution { publ…

Given inorder and postorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. 题目标签:Array, Tree 这到题目和105 几乎是一摸一样的,唯一的区别就是把pre-order 换成 post-order.因为post-order是最后一个数字是root,所以要从右向左的遍历.还需要把helpe…

Given inorder and postorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. 这个题目是给你一棵树的中序遍历和后序遍历,让你将这棵树表示出来.其中可以假设在树中没有重复的元素. 当做完这个题之后,建议去做做第105题,跟这道题类似. 分析:这个解法的基本思想是:我们有两个数组,分别是IN和POST.后…

我们都知道,已知中序和后序的序列是可以唯一确定一个二叉树的. 初始化时候二叉树为:================== 中序遍历序列,           ======O=========== 后序遍历序列,           =================O 红色部分是左子树,黑色部分是右子树,O是根节点 如上图所示,O是根节点,由后序遍历可知, 根据这个O可以把找到其在中序遍历当中的位置,进而,知道当前这个根节点O的左子树的前序遍历和中序遍历序列的范围. 以及右子树的前序遍历和中序遍历…

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. Show Tags Show Similar Problems 分析 跟上一道题同样的道理. AC代码 class Solution { public: template <typename Iter> TreeN…

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. 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 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&…

Given inorder and postorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. Subscribe to see which companies asked this question /** * Definition for a binary tree node. * struct TreeNode…

Problem Link: https://oj.leetcode.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/ This problem can be easily solved using recursive method. By given the inorder and postorder lists of the tree, i.e. inorder[1..n] and postorde…

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+后序遍历 ==== code: /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNod…

Given inorder and postorder traversal of a tree, construct the binary tree. 题目大意:给定一个二叉树的中序和后续序列,构建出这个二叉树. 解题思路:首先后序序列的最后一个是根节点,然后在中序序列中找到这个节点,中序序列中这个节点左边的是根节点的左子树,右边的是右子树,由此递归构建出完整的树. Talk is cheap: public TreeNode buildTree(int[] inorder, int[] pos…