1、preorder + inorder 第一个版本,使用坐标范围: 1/**2* Definition for binary tree3* struct TreeNode {4* int val;5* TreeNode *left;6* TreeNode *right;7* TreeNode(int x) : val(x), left(NULL), right(NULL) {}8* };9*/10classSolution {11public:12TreeNode *build(vector<int> ...
LeetCode: 889. Construct Binary Tree from Preorder and Postorder Traversal 题目描述 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 =...
1classSolution2{3public:4TreeNode* constructFromPrePost(vector<int>& pre, vector<int>&post)5{6if(pre.size()==0)7returnNULL;8TreeNode *node =newTreeNode(pre[0]);9if(pre.size()==1)10returnnode;11inti;12for(i =0;pre[1]!=post[i];i ++)13;14intlen = i+1;15//cout << len...
Construct Binary Tree from Preorder and Postorder Traversal // Runtime: 16 ms, faster than 40.63% of C++ online submissions for Construct Binary Tree from Preorder and Postorder Traversal. // Memory Usage: 15.1 MB, less than 16.67% of C++ online submissions for Construct Binary Tree from Pre...
而pre的意思就是说把中心放到了最前面,所以是ROOT, LEFT, RIGHT 而post就是说把中心放到了最后面,所以是LEFT,RIGHT,ROOT Construct Binary Tree from Inorder and Postorder Traversal 这道题就是给你两个数组,一个是Inorder traversal,一个是postorder,我之前做过类似的题目,但是还是想不出来,所以参考了自己以前...
Given inorder and postorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. 思路: easy 算法: 1. public TreeNode buildTree(int[] inorder, int[] postorder) { 2. if (inorder == null || inorder.length == 0 || postorde...
Given inorder and postorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. 翻译:给定树的中序遍历和后序遍历,构造二叉树。 注意:树中不存在重复项。 思路:本题与105. Construct Binary Tree from Preorder and Inorder Traversal类似。
1. Problem Descriptions:Given two integer arrays inorderandpostorderwhereinorderis the inorder traversal of a binary tree andpostorderis the postorder traversal of the same tree, construct and retu…
889. Construct Binary Tree from Preorder and Postorder Traversal,程序员大本营,技术文章内容聚合第一站。
Given inorder and postorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. 这道题要求从中序和后序遍历的结果来重建原二叉树,我们知道中序的遍历顺序是左-根-右,后序的顺序是左-右-根,对于这种树的重建一般都是采用递归来做,可参见...