DFS Recursion: public void DFS(TreeNode root){ if(root == null){ return; } System.out.println(root.val); DFS(root.left); DFS(root.right); } DFS Iteration: public void DFS(TreeNode root){ if(root == null){ return; } Stack<TreeNode> stk = new Stack<…

Given a binary tree, find its minimum depth. The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node. 题解:注意没有的分支不算. /** * Definition for a binary tree node. * struct TreeNode { * int val; * Tr…

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. /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left;…

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. 题意:给一棵二叉树,判断它是否是平衡二叉树(即每个节点的左右子树深度之…

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] ] 对每…

Given a binary tree, find the maximum path sum. The path may start and end at any node in the tree. For example:Given the below binary tree, 1 / \ 2 3   Return 6. 对于树,我们可以找到其左右子树中终止于根节点的最大路径值,称为最大半路径值,如果都是正数,则可以把它们以及根节点值相加,如果其中有负数,则舍弃那一边的值(即置为零).如此可以…

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. Hide Tags Tree Depth-first Search     简单的深度搜索 #include <iostream> using namespace std; /*…

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,…

这是悦乐书的第277次更新,第293篇原创 01 看题和准备 今天介绍的是LeetCode算法题中Easy级别的第145题(顺位题号是637).给定一个非空二叉树,以数组的形式返回每一层节点值之和的平均值.例如: 3 / \ 9 20 / \ 15 7 输出:[3,14.5,11] 说明:第一层上的节点的平均值为3,第二层上的节点的平均值为14.5,第三层上的节点的平均值为11.因此返回[3,14.5,11]. 注意:节点值的范围在32位有符号整数的范围内. 本次解题使用的开发工具是eclips…

Given a binary tree, find the maximum path sum. The path may start and end at any node in the tree. For example: Given the below binary tree, 1 / \ 2 3 Return 6. 解题思路: DFS暴力枚举,注意,如果采用static 全局变量的话,在IDE里面是可以通过,但在OJ上无法测试通过,因此需要建立一个类来储存结果,JAVA实现如下: publ…