leetcode------Binary Tree Level Order Traversal II

标题：	Binary Tree Level Order Traversal II
通过率：	30.5%
难度：	简单

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.

本题与前边我做的题目Binary Tree Level Order Traversal是一模一样的，第一个版本就是把树按层次进行输出，那么本题就是逆序，只用把最后的结果按照头插入法进行插入即可。

/**
 * Definition for binary tree
 * 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>> result=new LinkedList<List<Integer>>();
        LinkedList<TreeNode> queue=new LinkedList<TreeNode>();
        LinkedList<List<Integer>> ensort=new LinkedList<List<Integer>>();
        int count=1,level=0;
        if(root==null) return ensort;
        queue.addLast(root);
        while(!queue.isEmpty()){
            level=0;
            List<Integer> tmp=new ArrayList<Integer>();
            for(int i=0;i<count;i++){
                TreeNode tree=queue.removeFirst();
                tmp.add(tree.val);
                if(tree.left!=null){
                    queue.addLast(tree.left);
                    level++;
                }
                if(tree.right!=null){
                    queue.addLast(tree.right);
                    level++;
                }
            }
            ensort.addFirst(tmp);
            count=level;
        }
        //result=ensort;
        return ensort;
    }
}