Tree Representation Implementation & Traversal

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 right, level by level from leaf to root). For example:Given binary tree {3,9,20,#,#,15,7}, 3 / 9 20 / 15 7 re

[107-Binary Tree Level Order Traversal II(二叉树层序遍历II)] [LeetCode-面试算法经典-Java实现][全部题目文件夹索引] 原题 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

L102: 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 order

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#include<stdio.h> #include<queue> using namespace std; typedef struct BiTree { int val; struct BiTree *lchild; struct BiTree *rchild; }BiTree; void main() { BiTree *root; queue<BiTree *> q; BiTree *t; q.push(root); int size; //核心代码 while

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(); for(in

BFS以及它的扩展,我发现栈是个很好用的数据结构,特别是对于顺序需要颠倒的时候!!! 这里有个重要的信息:可以用null来标识一个level的结束!!! 下面是AC代码: 1 /** 2 * Given a binary tree, return the bottom-up level order traversal of its nodes' values. 3 * (ie, from left to right, level by level from leaf to root). 4 *

107. 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 right, level by level from leaf to root). For example:Given binary tree [3,9,20,null,null,15,7],     3  

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