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| 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 }
leetcode------Binary Tree Level Order Traversal II
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原文地址：http://www.cnblogs.com/pkuYang/p/4231608.html
