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.                                            …
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:…
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&…
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…
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…
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…
[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.…
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 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 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…
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…
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. 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. Subscribe to see which companies asked this question /** * Definition for a binary tree node. * struct TreeNode…
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…
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…
Given inorder and postorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. 给出二叉树的中序遍历和后序遍历结果,恢复出二叉树. 后序遍历序列的最后一个元素值是二叉树的根节点的值.查找该元素在中序遍历序列中的位置mid,依据中序遍历和后序遍历性质.有: 位置mid曾经的序列部分为二叉树根节点左子树中…
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 = [,,,,] postorder = [,,,,] Return the following binary tree: / \ / \ 中序.后序遍历得到二叉树,可以…
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. 题解: 这道题跟pre+in一样的方法做,只不过找左子树右子树的位置不同而已. / \ / \ / \ 对于上图的树来说, index: 0 1 2 3 4 5 6 中序遍历为为了清晰表示,我给节点上了颜色,红色是…
Given inorder and postorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. 解题思路: 给出一个二叉树的中序和后序遍历结果,还原这个二叉树. 对于一个二叉树: 1 / \ 2 3 / \ / \ 4 5 6 7 后序遍历结果为:4 5 2 6 7 3 1 中序遍历结果为:4 2 5 1 6 3 7…
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-&…
Given inorder and postorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree.同样是给出了不会有重复数字的条件,用递归较容易实现,代码如下: /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNod…
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. 说明: 1)实现与根据先序和中序遍历构造二叉树相似,题目参考请进 算法思想 中序序列:C.B.E.D.F.A.H.G.J.I   后序序列:C.E.F.D.B.H.J.I.G.A   递归思路: 根据后序遍历的特点,…
Given inorder and postorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. 题解:如下图所示的一棵树: 5 / \ 2 4 / \ \ 1 3 6 中序遍历序列:1  2  3  5  4  6 后序遍历序列:1  3  2  6  4  5 后序遍历序列的最后一个元素就是当前根节点元素.首先想到的…
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. 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…