*Binary Tree Level Order Traversal

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,#,#,15,7},

3 / \ 9 20 / \ 15 7

return its level order traversal as:

[ [3], [9,20], [15,7] ] 题解：这道题就是用传统的BFS来做。代码如下： public class Solution { public List<List<Integer>> levelOrder(TreeNode root) { List<List<Integer>> res = new LinkedList<List<Integer>>(); Queue<TreeNode> queue = new LinkedList<TreeNode>(); if(root==null) return res; queue.add(root); while(!queue.isEmpty()) { int levelnum = queue.size(); //这层有几个TreeNode List<Integer> sublist = new LinkedList<Integer>(); for(int i=0; i<levelnum;i++) { TreeNode node = queue.poll(); if(node.left!=null) queue.add(node.left); if(node.right!=null) queue.add(node.right); sublist.add(node.val); } res.add(sublist); } return res; } }

 

 

注意：

广度优先算法：用queue，先进先出。

入queue用add：Appends the specified element to the end of this list. This method is equivalent to addLast(E).

出queue用poll：Retrieves and removes the head (first element) of this list.

也可以用pop

 

深度优先算法：用stack，先进后出。

入stack用push：Pushes an element onto the stack represented by this list. In other words, inserts the element at the front of this list. This method is equivalent to addFirst(E).

出stack用pop：Pops an element from the stack represented by this list. In other words, removes and returns the first element of this list. This method is equivalent to removeFirst().

 

二者都可以用linkedlist定义，只是使用的时候methods不同，来决定是先进先出，还是先进后出。引用JDK API中关于LinkedList的一句说明："These operations allow linked lists to be used as a stack, queue, or double-ended queue."由此，可以得知，使用LinkedList可以轻松的实现栈和队列的功能。通过查看LinkedList源码实现，发现其底层采用双向链表实现。

LinkedList<TreeNode> queue = new LinkedList<TreeNode>();

reference:http://www.cnblogs.com/springfor/p/3891391.html

转载于:https://www.cnblogs.com/hygeia/p/4704027.html