构造方式跟中序与后序全然一样,并且一般都习惯正着来,所以更简单. 代码是之前写的,没实用库函数,不应该. TreeNode *buildIt(vector<int> &preorder, int start1, vector<int> &inorder, int start2, int len){ if(len <= 0) return NULL; TreeNode *root = new TreeNode(preorder[start1]); int i =…

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…

一天一道LeetCode 本系列文章已全部上传至我的github,地址:ZeeCoder's Github 欢迎大家关注我的新浪微博,我的新浪微博 欢迎转载,转载请注明出处 (一)题目 来源:https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/ Given preorder and inorder traversal of a tree, construct the bin…

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. 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.从前序以及中序的结果中构造二叉树,这里保证了不会有两个相同的数字,用递归构造就比较方便了: class Solution { public: TreeNode* buildTree(vector<int>& preor…

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…

这道题是为数不多的感觉在读本科的时候见过的问题. 人工构造的过程是如何呢.兴许遍历最后一个节点一定是整棵树的根节点.从中序遍历中查找到这个元素,就能够把树分为两颗子树,这个元素左側的递归构造左子树,右側的递归构造右子树.元素本身分配空间,作为根节点. 于set和map容器不同的是.vector容器不含find的成员函数.应该用stl的库函数,好在返回的也是迭代器,而vector的迭代器之间是能够做减法的.偏移量非常方便的得到. TreeNode *buildRec(vector<int> &a…

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.                                            …

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…