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. For example, given inorder = [,,,,] postorder = [,,,,] Return the following binary tree: / \ / \ 中序.后序遍历得到二叉树,可以…

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…

根据中序和后续遍历构建二叉树. /** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ class Solution { public TreeNode buildTree(int[] inorder, int[] postorder) { if(ino…

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root=1,然后我们在中序遍历中搜索1,可以看到中序遍历的第四个数是1,也就是root.根据中序遍历的定义,1左边的数{4,2,5}就是左子树的中序遍历,1右边的数{6,3,7}就是右子树的中序遍历…

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 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 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.…

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…

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…

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…

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 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. 题意:根据中序遍历和后序遍历,构建二叉树 思路很清晰,做法很简单,就不讲了. 一开始我写了一个递归的解法,本地测试数据都OK,无奈提交的时候内存超出限制,下面先给出超出内存的代码: /** * Definition for…

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…

［抄题］: 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 /…

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…

题目: Given inorder and postorder traversal of a tree, construct the binary tree. 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,右边就是右子树的中序,然后根据刚才中序序列中左右子树的元素个数可以在后序序列中找到左右子树的后序序列,然后递归的求解即可. /** * Definition for a binar…