Java [Leetcode 107]Binary Tree Level Order Traversal II

题目描述：

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 traversal as:

[
  [15,7],
  [9,20],
  [3]
]

解题思路：

运用广度优先搜索方法，运用队列的方法，每次遍历一层的叶节点，并把下一层的叶节点加入到队列中。每次遍历一层节点结束时候，将该层节点组成的list放到整体的list之前，实现由底到高的排列。

代码如下：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<List<Integer>> levelOrderBottom(TreeNode root) {
    	List<List<Integer>> list = new LinkedList<List<Integer>>();
    	Queue<TreeNode> queue = new LinkedList<TreeNode>();
    	if(root == null)
    		return list;
    	queue.offer(root);
    	while(!queue.isEmpty()){
    		int num = queue.size();
    		List<Integer> levelList = new LinkedList<Integer>();
    		for(int i = 0; i < num; i++){
    			if(queue.peek().left != null)
    				queue.offer(queue.peek().left);
    			if(queue.peek().right != null)
    				queue.offer(queue.peek().right);
    			levelList.add(queue.poll().val);
    		}
    		list.add(0, levelList);
    	}
    	return list;
    }
}