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 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] ] 分层…

Binary Tree Level Order Traversal 题目描述: 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…

Binary Tree Level Order Traversal OJ: https://oj.leetcode.com/problems/binary-tree-level-order-traversal/ 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…

LeetCode:Binary Tree Level Order Traversal 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 ord…

题目: 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 orde…

Binary Tree Level Order Traversal 二叉树的层序遍历 两种方式: 1.用两个queue交替表示每一层的节点 2.用两个node,一个表示当前层的最后一个节点,一个表示下一层的最后一个节点 /** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x…

思路: 与Binary Tree Level Order Traversal I 几乎一样.只是最后将结果存放在栈里,然后在栈里再传给向量即可. 再次总结思路: 两个queue,先把第一个放进q1,循环q1是否为空,不为空就读取并出列,如果root有孩子就放入q2,最后清空q2. 注意: for循环的时候不要使用vector.size()这类作为最大值判断,由于vector的size可能不断的减小,这回导致遍历不完的情况发生. 即: int count = sret.size(); ;i< co…

Question 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…