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. 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. 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…

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…

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 <=…

原题 题意: 根据先序和中序得到二叉树(假设无重复数字) 思路: 先手写一次转换过程,得到思路. 即从先序中遍历每个元素,(创建一个全局索引,指向当前遍历到的元素)在中序中找到该元素作为当前的root,以该节点左边所有元素作为当前root的左支,右同理. 重复分别对左右边所有元素做相同处理. class Solution { public: TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder…

不用迭代器的代码 class Solution { public: TreeNode* reConstructBinaryTree(vector<int> pre,vector<int> vin) { TreeNode* root = NULL; int length_pre = pre.size(); int length_vin = vin.size(); || length_vin <= ) return root; ,length_pre-,,length_vin-)…

原题地址 基本二叉树操作. O[       ][              ] [       ]O[              ] 代码: TreeNode *restore(vector<int> &preorder, vector<int> &inorder, int pp, int ip, int len) { ) return NULL; TreeNode *node = new TreeNode(preorder[pp]); ) return node…