leetcode之Construct Binary Tree from Preorder and Inorder Traversal

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(

问题: 给定二叉树的前序和中序遍历,重构这课二叉树. 分析: 前序.中序.后序都是针对于根结点而言,所以又叫作先根.中根.后根(当然不是高跟). 前序:根  左 右 中序:左  根 右 对二叉树,我们将其进行投影,就会发现个有趣的事: 发现投影后的顺序,恰好是中序遍历的顺序,这也就是为什么在构造二叉树的时候,一定需要知道中序遍历,因为中序遍历决定了结点间的相对左右位置关系.所以,对一串有序的数组,我们可以来构建二叉有序数,并通过中序遍历,就可以得到这个有序的数组. 既然中序遍历可以通过根结点将序

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. 思路 主要是根据前序遍历和中序遍历的特点解决这个题目. 1.确定树的根节点.树根是当前树中所有元素在前序遍历中最先出现的元素. 2.求解树的子树.找出根节点在中序遍历中的位置,根左边的所有元素就是左子树,根右边的所有元

Construct Binary Tree from Preorder and Inorder Traversal Total Accepted: 14824 Total Submissions: 55882My Submissions Given preorder and inorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the

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. 1 / \ 2        

Given preorder and inorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. 思路:首先根据前序遍历得到根节点,然后在中序遍历中得到根节点的位置,左边的为左子树,右边的为右子树. 然后再递归求解左子树和右子树的构造即可.代码如下: /** * Definition for a binary tree

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; *

Question: Given preorder and inorder traversal of a tree, construct the binary tree. 根据树的前序遍历和中序遍历,构建二叉树 Algorithm: 前序遍历:根-左-右 中序遍历:左-根-右 举个例子 前序遍历:ABDECFG 中序遍历:DBEAFCG 1.前序遍历的第一个元素A,为根结点 2.在中序遍历找到A,则在中序遍历中A左边是左子树,右边是右子树 3.递归(左右子树同样如此) Accepted Code:

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