LeetCode-Binary Tree Level Order Traversal II

Binary Tree Level Order Traversal II

##题目

####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

return its bottom-up level order traversal as:

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

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

####OJ’s Binary Tree Serialization:
The serialization of a binary tree follows a level order traversal, where ‘#’ signifies a path terminator where no node exists below.

Here’s an example: 

  1
 / \
2   3
   /
  4
   \
    5

The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".

##解题思路
该题和Binary Tree Level Order Traversal的解法相同，只不过在最后结果上是其结果的转置。

这道题我没有想到什么好的做法可以一次的自底向上进行层序遍历，能想到的就是进行Binary Tree Level Order Traversal中的遍历，然后对结果进行一次reverse.

##算法代码
代码采用JAVA实现：

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/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ArrayList<ArrayList<Integer>> levelOrderBottom(TreeNode root) {
    	ArrayList<ArrayList<Integer>> res=new ArrayList<ArrayList<Integer>>();
    	if(root==null)
    		return res;
    	//使用队列来存储节点
    	LinkedList<TreeNode> queen=new LinkedList<TreeNode>();
    	//使用队列来存储上述队列中节点的层数
    	LinkedList<Integer> cequeen=new LinkedList<Integer>();
    	queen.addLast(root);
    	cequeen.addLast(0);
    	int ce=0;
    	ArrayList<Integer> cur=new ArrayList<Integer>();
    	while(!queen.isEmpty())
    	{
    		if(ce==cequeen.getFirst())
    		{
    			TreeNode no=queen.removeFirst();
    			int cengshu=cequeen.removeFirst();
    			cur.add(no.val);
    			if(no.left!=null)
    			{
    				queen.addLast(no.left);
    				cequeen.addLast(cengshu+1);
    			}
    			if(no.right!=null)
    			{
    				queen.addLast(no.right);
    				cequeen.addLast(cengshu+1);
    			}
    		}else{
    			res.add(new ArrayList<Integer>(cur));
    			ce++;
    			cur.clear();
    		}
    	}
    	//最后一次不要忘记加入
    	res.add(new ArrayList<Integer>(cur));
    	Collections.reverse(res);
    	return res;

    }
}

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