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…

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. 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…

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. 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…

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…

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…

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: /** *…

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: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]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. 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 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…

Return any binary tree that matches the given preorder and postorder traversals. Values in the traversals pre and post are distinct positive integers. Example 1: Input: pre = [1,2,4,5,3,6,7], post = [4,5,2,6,7,3,1] Output: [1,2,3,4,5,6,7] Note: 1 <=…

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…

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.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…

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…

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. 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. 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. 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. 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…