Given two binary strings, return their sum (also a binary string). For example,a = "11"b = "1"Return "100". public class Solution { public String addBinary(String a, String b) { if(a.equalsIgnoreCase("")||a==null) r
Given a binary tree, determine if it is a valid binary search tree (BST). Assume a BST is defined as follows: The left subtree of a node contains only nodes with keys less than the node's key. The right subtree of a node contains only nodes with keys
转载请注明出处:http://blog.csdn.net/crazy1235/article/details/51524241 Subject 出处:https://leetcode.com/problems/binary-tree-zigzag-level-order-traversal/ Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to ri
[题目] 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. [说明] 不是很难,思路大家可能都会想到用递归,分别判断左右
转载请注明出处:http://blog.csdn.net/crazy1235/article/details/51494797 Subject 出处:https://leetcode.com/problems/binary-tree-postorder-traversal/ Hard 级别 Given a binary tree, return the postorder traversal of its nodes' values. For example: Given binary tree
转载请注明出处:http://blog.csdn.net/crazy1235/article/details/51508308 Subject 出处:https://leetcode.com/problems/binary-tree-level-order-traversal-ii/ Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to rig
[题目] Two elements of a binary search tree (BST) are swapped by mistake. Recover the tree without changing its structure. Note: A solution using O(n) space is pretty straight forward. Could you devise a constant space solution? [解析] 题意:二叉搜索树中,有两个结点的位置
[题目] Given inorder and postorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. [解析] 题意:根据二叉树中序遍历和后序遍历的结果,构造该二叉树. 首先明确一下,中序遍历顺序:left - root - right,后序遍历顺序:left - right - root. 很显然,后序遍历的