Binary Tree Traversals Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3475    Accepted Submission(s): 1555 Problem Description A binary tree is a finite set of vertices that is either empty or…

Binary Tree Traversals Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2442    Accepted Submission(s): 1063 Problem Description A binary tree is a finite set of vertices that is either empty or…

                                                                            Binary Tree Traversals Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Description A binary tree is a finite set of vertices that i…

传送门 Description A binary tree is a finite set of vertices that is either empty or consists of a root r and two disjoint binary trees called the left and right subtrees. There are three most important ways in which the vertices of a binary tree can be…

A binary tree is a finite set of vertices that is either empty or consists of a root r and two disjoint binary trees called the left and right subtrees. There are three most important ways in which the vertices of a binary tree can be systematically…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1710 题意:给前序.中序求后序,多组 前序:根左右 中序:左右根 分析:因为前序(根左右)最先出现的总是根结点,所以令root为前序中当前的根结点下标(并且同时把一棵树分为左子树和右子树).start为当前需要打印的子树在中序中的最左边的下标,end为当前需要打印的子树在中序中最右边的下标.递归打印这棵树的后序,递归出口为start > end.i为root所表示的值在中序中的下标,所以i即是分隔中…

Binary Tree Traversals Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2681    Accepted Submission(s): 1178 Problem Description A binary tree is a finite set of vertices that is either empty or…

Binary Tree Traversals Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4330    Accepted Submission(s): 1970 Problem Description A binary tree is a finite set of vertices that is either empty or…

题目地址:HDU 1710 已知二叉树先序和中序求后序. #include <stdio.h> #include <string.h> int a[1001], cnt; typedef struct node { int date ; node *lchild , *rchild ; }*tree; int getk(int ch,int ino[],int is,int n) { for(int i = is ; i <= is + n -1 ; i++) if(ino[…

http://acm.hdu.edu.cn/showproblem.php?pid=1710 已知先序和中序遍历,求后序遍历二叉树. 思路:先递归建树的过程,后后序遍历. Binary Tree Traversals Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3442    Accepted Submission(s): 1541…

http://acm.hdu.edu.cn/showproblem.php?pid=1710 Binary Tree Traversals Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4210    Accepted Submission(s): 1908 Problem Description A binary tree is a…

Problem Description         A binary tree is a finite set of vertices that is either empty or consists of a root r and two disjoint binary trees called the left and right subtrees. There are three most important ways in which the vertices of a binary…

Binary Tree Traversals Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 9283    Accepted Submission(s): 4193 Problem Description A binary tree is a finite set of vertices that is either empty or…

题链;http://acm.hdu.edu.cn/showproblem.php?pid=1710 Binary Tree Traversals Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4205    Accepted Submission(s): 1904 Problem Description A binary tree i…

Given a binary tree, return the preorder traversal of its nodes' values. For example:Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,2,3]. Note: Recursive solution is trivial, could you do it iteratively? 树的遍历,最常见的有先序遍历,中序遍历,后序遍历和层序遍历,它们用递归实现起来都非常的简…

1.HDU  1710  Binary Tree Traversals 2.链接:http://acm.hust.edu.cn/vjudge/problem/33792 3.总结:记录下根结点,再拆分左右子树,一直搜下去.感觉像dfs. 题意:二叉树,输入前.中序求后序. (1)建立出一颗二叉树,更直观.但那些指针用法并不是很懂. #include<iostream> #include<cstdio> #include<cstdlib> using namespace…

Tree Recovery Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 14640   Accepted: 9091 Description Little Valentine liked playing with binary trees very much. Her favorite game was constructing randomly looking binary trees with capital le…

题目链接 题目大意: 输入二叉树的前序.中序遍历,请输出它的后序遍历 #include <stdio.h> #include <string.h> ; // 长度为n s1 前序 s2 中序 构造后序s3 void build(int n, char * s1, char * s2, char * s3) { ) return; ]) - s2; //找到根节点在中序遍历中的位置 build(p, s1 + , s2, s3); //递归左子树的后序遍历 build(n - p -…

acm.hdu.edu.cn/showproblem.php?pid=1710 [题意] 给定一棵二叉树的前序遍历和中序遍历,输出后序遍历 [思路] 根据前序遍历和中序遍历递归建树,再后续遍历输出 malloc申请空间在堆,函数返回,内存不释放,需要free手动释放 [Accepted] #include<iostream> #include<cstdio> #include<string> #include<cstring> #include<alg…

Given a binary tree where all the right nodes are either leaf nodes with a sibling (a left node that shares the same parent node) or empty, flip it upside down and turn it into a tree where the original right nodes turned into left leaf nodes. Return…

Given a binary tree, return the postorder traversal of its nodes' values. For example: Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [3,2,1]. Note: Recursive solution is trivial, could you do it iteratively? 经典题目,求二叉树的后序遍历的非递归方法,跟前序,中序,层序一样都需要用到栈,后续的…

Given a binary tree, return the inorder traversal of its nodes' values. For example:Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,3,2]. Note: Recursive solution is trivial, could you do it iteratively? confused what "{1,#,2,3}" means? > re…

题目: 二叉树的后序遍历 给出一棵二叉树,返回其节点值的后序遍历. 样例 给出一棵二叉树 {1,#,2,3}, 1 \ 2 / 3 返回 [3,2,1] 挑战 你能使用非递归实现么? 解题: 递归程序好简单 Java程序: /** * Definition of TreeNode: * public class TreeNode { * public int val; * public TreeNode left, right; * public TreeNode(int val) { * th…

题目: 二叉树的前序遍历 给出一棵二叉树,返回其节点值的前序遍历. 样例 给出一棵二叉树 {1,#,2,3}, 1 \ 2 / 3 返回 [1,2,3]. 挑战 你能使用非递归实现么? 解题: 通过递归实现,根节点->左节点->右节点 Java程序: /** * Definition of TreeNode: * public class TreeNode { * public int val; * public TreeNode left, right; * public TreeNode(…

题目意思:二叉树中序遍历,结果存在vector<int>中 解题思路:迭代 迭代实现: /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: vector&l…

题目意思:二叉树先序遍历,结果存在vector<int>中 解题思路:1.递归(题目中说用递归做没什么意义,我也就贴贴代码吧) 2.迭代 迭代实现: class Solution { public: vector<int> preorderTraversal(TreeNode* root) { vector<int> ans; if(root){ TreeNode* temp; stack<TreeNode*> s; //利用栈,每次打印栈顶,然后将栈顶弹出…

Given a binary tree, return the tilt of the whole tree. The tilt of a tree node is defined as the absolute difference between the sum of all left subtree node values and the sum of all right subtree node values. Null node has tilt 0. The tilt of the …

Given a binary tree, return all root-to-leaf paths. For example, given the following binary tree: 1 / \ 2 3 \ 5 All root-to-leaf paths are: ["1->2->5", "1->3"] 题目标签:Tree 这道题目给了我们一个二叉树,让我们记录所有的路径,返回一个array string list. 我们可以另外设一…

Invert a binary tree. 4 / \ 2 7 / \ / \ 1 3 6 9 to 4 / \ 7 2 / \ / \ 9 6 3 1 Trivia:This problem was inspired by this original tweet by Max Howell: Google: 90% of our engineers use the software you wrote (Homebrew), but you can’t invert 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. 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…