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。

 1 /**
 2  * Definition for binary tree
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public List<List<Integer>> levelOrderBottom(TreeNode root) {
12          //List<List<Integer>> result=new LinkedList<List<Integer>>();
13         LinkedList<TreeNode> queue=new LinkedList<TreeNode>();
14         LinkedList<List<Integer>> ensort=new LinkedList<List<Integer>>();
15         int count=1,level=0;
16         if(root==null) return ensort;
17         queue.addLast(root);
18         while(!queue.isEmpty()){
19             level=0;
20             List<Integer> tmp=new ArrayList<Integer>();
21             for(int i=0;i<count;i++){
22                 TreeNode tree=queue.removeFirst();
23                 tmp.add(tree.val);
24                 if(tree.left!=null){
25                     queue.addLast(tree.left);
26                     level++;
27                 }
28                 if(tree.right!=null){
29                     queue.addLast(tree.right);
30                     level++;
31                 }
32             }
33             ensort.addFirst(tmp);
34             count=level;
35         }
36         //result=ensort;
37         return ensort;
38     }
39 }

转载于:https://www.cnblogs.com/pkuYang/p/4231608.html

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