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: Just do it recursively. TreeNode *build(vector<int> &preorder, int pstart, int pend, vector<int…

Given preorder and inorder traversal of a tree, construct the binary tree. 题目大意:给定一个二叉树的前序和中序序列,构建出这个二叉树. 解题思路:跟中序和后序一样,在先序中取出根节点,然后在中序中找到根节点,划分左右子树,递归构建这棵树,退出条件就是左右子树长度为1,则返回这个节点,长度为0则返回null. Talk is cheap: public TreeNode buildTree(int[] preorder,…

题目: Given preorder and inorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. 题解: / \ / \ / \ 对于上图的树来说, index: 0 1 2 3 4 5 6 先序遍历为                                                        …

Construct Binary Tree from Preorder and Inorder Traversal Given preorder and inorder traversal of a tree, construct the binary tree. 分析: 根据前序遍历和中序遍历构造一棵树,递归求解即可 /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left;…

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. SOLUTION 1: 1. Find the root node from the preorder.(it…

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. 可与Construct Binary Tree from Inorder and Postorder Trav…

Given preorder and inorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. 这道题要求用先序和中序遍历来建立二叉树,跟之前那道Construct Binary Tree from Inorder and Postorder Traversal 由中序和后序遍历建立二叉树原理基本相同,针对这道题,由于先…

Given preorder and inorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. For example, given preorder = [3,9,20,15,7] inorder = [9,3,15,20,7] Return the following binary tree: 3 / \ 9 20…

Problem Link: https://oj.leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/ The basic idea is same to that for Construct Binary Tree from Inorder and Postorder Traversal. We solve it using a recursive function. First, we…

Given preorder and inorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. 此题目有两种解决思路: 1)递归解决(比较好想)按照手动模拟的思路即可 2)非递归解决,用stack模拟递归 class Solution { public: TreeNode *buildTree(vector<int>&…

题目: Given preorder and inorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. 解题思路分析: 前序遍历首先遍历根节点,然后依次遍历左右子树 中序遍历首先遍历左子树,然后遍历根结点,最后遍历右子树 根据二者的特点,前序遍历的首节点就是根结点,然后在中序序列中找出该结点位置(index),则该结点之…

Given preorder and inorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. 给出前序遍历和中序遍历,然后求这棵树. 很有规律.递归就可以实现. /** * Definition for a binary tree node. * public class TreeNode { * int val; *…

Given preorder and inorder traversal of a tree, construct the binary tree. ============== 基本功: 利用前序和中序构建二叉树 , === code /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : va…

题目: Given preorder and inorder 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 *rig…

原题地址: https://oj.leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/ 题目内容: Given preorder and inorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. 方法: 非常单纯的…

Given preorder and inorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. 通过树的中序序列和前序序列来构建一棵树. 例如:一棵树的前序序列是1-2-3,中序序列有5种可能,分别是1-2-3.2-1-3.1-3-2.2-3-1.3-2-1.不同的中序序列对应着不同的树的结构,这几棵树分别是[1,nul…

Given preorder and inorder 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     递归就可以了.   #include<algorithm> using namespace std; /**…

Given preorder and inorder 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 题目给了我们preOrder 和 inOrder 两个遍历array,让我们建立二叉树.先来举一个例子,让我们看一下preOrder 和 inOrder的特性. / \   /      \…

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Given preorder and inorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. For example, given preorder = [3,9,20,15,7] inorder = [9,3,15,20,7] Return the following binary tree: 3 / \ 9 20…

Given preorder and inorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. For example, given preorder = [3,9,20,15,7] inorder = [9,3,15,20,7] Return the following binary tree: 3 / \ 9 20…

Given preorder and inorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. For example, given preorder = [,,,,] inorder = [,,,,] Return the following binary tree: / \ / \ 前序.中序遍历得到二叉树,可以知道…

［抄题］: Given preorder and inorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. For example, given preorder = [3,9,20,15,7] inorder = [9,3,15,20,7] Return the following binary tree: 3 / \…

题目: Given preorder and inorder traversal of a tree, construct the binary tree. 思路: 线序序列的第一个元素就是树根,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 binary…

Given preorder and inorder 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> &preorder, vector<int> &inorder) { root…

Given preorder and inorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. 解题思路: 给出一个二叉树的先序和中序遍历结果,还原这个二叉树. 对于一个二叉树: 1 / \ 2 3 / \ / \ 4 5 6 7 先序遍历结果为:1 2 4 5 3 6 7 中序遍历结果为:4 2 5 1 6 3 7 由…

Given preorder and inorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. For example, given preorder = [3,9,20,15,7] inorder = [9,3,15,20,7] Return the following binary tree: 3 / \ 9 20…