Problem Link: http://oj.leetcode.com/problems/binary-tree-postorder-traversal/ The post-order-traversal of a binary tree is a classic problem, the recursive way to solve it is really straightforward, the pseudo-code is as follows. RECURSIVE-POST-ORDE…

1 题目 Given two binary strings, return their sum (also a binary string). For example,a = "11"b = "1"Return "100". 接口 String addBinary(String a, String b) 2 思路 处理二进制求和和进位.从低位开始,一直相加并且维护进位.和Add Two Numbers的区别是这个题目低位在后面,所以要从strin…

Problem Link: https://oj.leetcode.com/problems/binary-tree-level-order-traversal/ Traverse the tree level by level using BFS method. # Definition for a binary tree node # class TreeNode: # def __init__(self, x): # self.val = x # self.left = None # se…

Problem Link: https://oj.leetcode.com/problems/binary-tree-zigzag-level-order-traversal/ Just BFS from the root and for each level insert a list of values into the result. # Definition for a binary tree node # class TreeNode: # def __init__(self, x):…

Problem Link: https://oj.leetcode.com/problems/binary-tree-level-order-traversal-ii/ Use BFS from the tree root to traverse the tree level by level. The python code is as follows. # Definition for a binary tree node # class TreeNode: # def __init__(s…

Problem Link: http://oj.leetcode.com/problems/binary-tree-maximum-path-sum/ For any path P in a binary tree, there must exists a node N in P such that N is the ancestor node of all other nodes in P. We call such N as the root of P, or P roots at N. T…

Problem Link: http://oj.leetcode.com/problems/binary-tree-preorder-traversal/ Even iterative solution is easy, just use a stack storing the nodes not visited. Each iteration, pop a node and visited it, then push its right child and then left child in…

Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root). For example:Given binary tree {3,9,20,#,#,15,7}, 3 / \ 9 20 / \ 15 7 return its bottom-up level order tr…

http://oj.leetcode.com/problems/validate-binary-search-tree/ 判断一棵树是否为二叉搜索树.key 是,在左子树往下搜索的时候,要判断是不是子树的值都小于跟的值,在右子树往下搜索的时候,要判断,是不是都大于跟的值.很好的一个递归改进算法. 简洁有思想! #include<climits> // Definition for binary tree struct TreeNode { int val; TreeNode *left; Tr…

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"] Credits:Special thanks to @jianchao.li.fighter for adding this pro…

Given a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom. For example:Given the following binary tree, 1 <--- / \ 2 3 <--- \ \ 5 4 <--- You should return [1, 3,…

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?中序遍历二叉树,递归遍历当然很容易,题目还要求不用递归,下面给出两种方法: 递归: /**…

Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between). For example:Given binary tree {3,9,20,#,#,15,7}, 3 / \ 9 20 / \ 15 7 return its…

Given a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom. For example:Given the following binary tree, 1 <--- / \ 2 3 <--- \ \ 5 4 <--- You should return [1, 3,…

Implement an iterator over a binary search tree (BST). Your iterator will be initialized with the root node of a BST. Calling next() will return the next smallest number in the BST Note: next() and hasNext() should run in average O(1) time and uses O…

Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root). For example:Given binary tree {3,9,20,#,#,15,7}, 3 / \ 9 20 / \ 15 7 return its bottom-up level order tr…

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}, 3 / \ 9 20 / \ 15 7 return its level order traversal as: [ [3], [9,20], [15,7] ] 层序…

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"] 简单的遍历查找路径问题,代码如下: /** * Definition for a binary tree node. * struc…

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]. 前序遍历二叉树,只不过题目的要求是尽量不要使用递归,当然我还是先用递归写了一个: /** * Definition for a binary tree node. * struct TreeNode { * int val…

http://oj.leetcode.com/problems/unique-binary-search-trees-ii/ 一题要求得出所有树的种类数,二题要求得出所有树. 在一题的基础上修改代码,还是要提前想清楚再写. #include <iostream> #include <vector> using namespace std; struct TreeNode { int val; TreeNode *left; TreeNode *right; TreeNode(int…

Given two binary strings, return their sum (also a binary string). For example,a = "11"b = "1"Return "100". 题目标签:Math 题目给了我们两个string a 和 b,让我们把这两个二进制 相加. 首先把两个string 的长度得到,然后从右向左 取 两个string 的 digit. 增设一个 carry = 0: 每一轮把 digit…

Given two binary strings, return their sum (also a binary string). The input strings are both non-empty and contains only characters 1 or 0. Example 1: Input: a = "11", b = "1" Output: "100" Example 2: Input: a = "1010&q…

题目简述: Given two binary strings, return their sum (also a binary string). For example, a = "11" b = "1" Return "100". 解题思路: class Solution: # @param a, a string # @param b, a string # @return a string def addBinary(self, a, b)…

Given two binary strings, return their sum (also a binary string). For example, a = "11" b = "1" Return "100". 解题思路: JAVA实现如下: static public String addBinary(String a, String b) { if (a.length() < b.length()) { String temp…

Given two binary strings, return their sum (also a binary string). For example,a = "11"b = "1"Return "100". 这个题目只要注意各种情况你就成功了一大半,特别要注意的是对进位赋值后可能产生的变化,以及最后一位进位为1时, 我们要把这个1插进来.同时注意字符串的低位是我们数值的高位 class Solution { public: string…

题目描述: Given two binary strings, return their sum (also a binary string). For example,a = "11"b = "1"Return "100". 解题思路: 使用StringBuilder,且使用进位. 代码如下: public class Solution { public String addBinary(String a, String b) { String…

题目: Given two binary strings, return their sum (also a binary string). For example, a = "11" b = "1" Return "100". 思路: 题意:对字符串的二进制数字,计算 把二进制转化为整数,设置变量carry位进位,sum%2,是相应的数字,carry = sum/2 - 代码: public class Solution { public St…

Given two binary strings, return their sum (also a binary string). The input strings are both non-empty and contains only characters 1or 0. Example 1: Input: a = "11", b = "1" Output: "100" Example 2: Input: a = "1010&qu…

题目 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,null,null,15,7], 3 / \ 9 20 / \ 15 7 return its level order traversal as: [ [3], [9,20], […

Given two binary strings, return their sum (also a binary string). The input strings are both non-empty and contains only characters 1 or 0. Example 1: Input: a = "11", b = "1" Output: "100" Example 2: Input: a = "1010&q…