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…

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. 说明: 1)二叉树可空 2)思路:a.根据前序遍历的特点, 知前序序列(PreSequence)的首个元素(PreSequence[0])为二叉树的根(root),  然后在中序序列(InSequence)中查找此根(…

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…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 递归 日期 题目地址:https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/description/ 题目描述 Given preorder and inorder traversal of a tree, construct t…

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. 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. / \   /      \…

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. Note: You may assume that duplicates do not exist in the tree. 解题思路一: preorder[0]为root,以此分别划分出inorderLeft.preorderLeft.inorderRight.preorderRight四个数组,然后root.left=buildTree(pre…

Given preorder and inorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. public class Solution { public TreeNode buildTree(int[] preorder, int[] inorder) { if(preorder.length==0||inorder…

根据前序中序构建二叉树. 1 / \ 2 3 / \ / \ 4 5 6 7对于上图的树来说, index: 0 1 2 3 4 5 6 先序遍历为: 6 3 7为了清晰表示,我给节点上了颜色,红色是根节点,蓝色为左子树,绿色为右子树.可以发现的规律是:1. 先序遍历的从左数第一个为整棵树的根节点.2. 中序遍历中根节点是左子树右子树的分割点.再看这个树的左子树: 先序遍历为: 2 4 5 中序遍历为: 4 2 5依然可以套用上面发现的规律.右子树: 先序遍历为: 3 6 7 中序遍历为: 6…