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

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.…

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. 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…

原题地址 思路: 和leetcode105题差不多,这道题是给中序和后序,求出二叉树. 解法一: 思路和105题差不多,只是pos是从后往前遍历,生成树顺序也是先右后左. class Solution { public: TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) { int pos = postorder.size() - 1; return dfs(inorder, post…

原题地址 二叉树基本操作 [       ]O[              ] [       ][              ]O 代码: TreeNode *restore(vector<int> &inorder, vector<int> &postorder, int ip, int pp, int len) { ) return NULL; TreeNode *node = ]); ) return node; ; ]) leftLen++; node-&…

[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 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…

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. 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. 和上一道题基本一样 根据中序和后序遍历确定一个棵树. /** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeN…

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

［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 /…

Given inorder and postorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. struct TreeNode { int val; TreeNode *left; TreeNode *right; TreeNode(int x) : val(x), left(NULL), right(NULL) {…

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…

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…

给定一棵树的中序遍历与后序遍历,依据此构造二叉树.注意:你可以假设树中没有重复的元素.例如,给出中序遍历 = [9,3,15,20,7]后序遍历 = [9,15,7,20,3]返回如下的二叉树:    3   / \  9  20    /  \   15   7详见:https://leetcode.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/description/ Java实现: /** *…

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…

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…

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:…

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…

原题链接在这里:https://leetcode.com/problems/construct-binary-tree-from-preorder-and-postorder-traversal/ 题目: Return any binary tree that matches the given preorder and postorder traversals. Values in the traversals pre and post are distinct positive intege…