题意:二叉树的根节点为(1,1),对每个结点(a,b)其左结点为 (a + b, b) ,其右结点为 (a, a + b),已知某结点坐标,求根节点到该节点向左和向右走的次数. 分析:往回一步一步走肯定超时,不过如果a>b,大的有点多,就没必要一步一步走,直接a/b就好了,然后把a变成a%b取余,那么a/b就是从现在的a到原来的a向左走的步数.(a<b同理) 同理,如果a=1的话,那么也没必要往回走了,因为从根节点到现在的结点就是向右走了b-1.(b=1同理) #pragma comment(…
Given a binary tree, determine if it is height-balanced. For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1. 递归,求左右是否平衡,再判断高度差 /** * Definition…
题目地址:http://poj.org/problem?id=1032 Parliament Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 17473 Accepted: 7371 Description New convocation of The Fool Land's Parliament consists of N delegates. According to the present regulation…
Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems. You are given a tree with nn nodes. In the beginning, 00 is written on all edges. In one operation, you can choose any 22 dist…
UVA 1394 And Then There Was One / Gym 101415A And Then There Was One / UVAlive 3882 And Then There Was One / POJ 3517 And Then There Was One / Aizu 1275 And Then There Was One (动态规划,思维题) Description Let's play a stone removing game. Initially, n ston…
既上篇关于二叉搜索树的文章后,这篇文章介绍一种针对二叉树的新的中序遍历方式,它的特点是不需要递归或者使用栈,而是纯粹使用循环的方式,完成中序遍历. 线索二叉树介绍 首先我们引入“线索二叉树”的概念: "A binary tree is threaded by making all right child pointers that would normally be null point to the inorder successor of the node, and all left chi…
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? 解题:果然不能晚上做题,效率好低.看了讨论才学会的解法.设置一个指针next指向当前访问的…
Binary Tree Preorder Traversal 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? ''' Created o…
Binary Tree Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 6355 Accepted: 2922 Description Background Binary trees are a common data structure in computer science. In this problem we will look at an infinite binary tree where the node…
Given inorder and postorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. 给出二叉树的中序遍历和后序遍历结果,恢复出二叉树. 后序遍历序列的最后一个元素值是二叉树的根节点的值.查找该元素在中序遍历序列中的位置mid,依据中序遍历和后序遍历性质.有: 位置mid曾经的序列部分为二叉树根节点左子树中…
Binary Tree Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 138 Accepted Submission(s): 73 Special Judge Problem Description The Old Frog King lives on the root of an infinite tree. Accordin…
题目链接 Problem Description You are given a complete binary tree with n nodes. The root node is numbered 1, and node x's father node is ⌊x/2⌋. At the beginning, node x has a value of exactly x. We define the value of a path as the sum of all nodes it pa…
Given a binary tree, find its maximum depth. The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node. 解题: 递归就行:根节点为空,返回0:一个子树根节点为空,另一个子树根节点不为空,就返回根节点不为空的子树高度:否则返回两个子树中高度大者加一. 代码: /** * Definit…
Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level). For example:Given binary tree {3,9,20,#,#,15,7}, / \ / \ return its level order traversal as: [ [], [,], [,] ] 题解:二叉树的层次遍历,用队列实现.重点在…
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? 解题:应该是很简单的一道题,纠结了好久T_T 基本思路很简单,用栈模拟就可以了.首先根节…
Given preorder and inorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. 类似http://www.cnblogs.com/sunshineatnoon/p/3854935.html 只是子树的前序和中序遍历序列分别更新为: //左子树: left_prestart = prestart+1 lef…
