Leet Code OJ 107. Binary Tree Level Order Traversal II [Difficulty: Easy]
题目：
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},
return its bottom-up level order traversal as:
[
[15,7],
[9,20],
[3]
]
翻译：
给定一个二叉树，返回它的节点的从底部到头部的层序遍历结果。
代码：
/*** 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>> result=levelOrder(root);Collections.reverse(result);return result;}public List<List<Integer>> levelOrder(TreeNode root) {List<List<Integer>> result=new ArrayList<>();if(root==null){return result;}List<Integer> one=new ArrayList<>();one.add(root.val);result.add(one);List<List<Integer>> left=levelOrder(root.left);List<List<Integer>> right=levelOrder(root.right);for(int i=0;i<left.size()||i<right.size();i++){List<Integer> item=new ArrayList<>();if(i<left.size()){for(Integer k:left.get(i)){item.add(k);}}if(i<right.size()){for(Integer k:right.get(i)){item.add(k);}}result.add(item);}return result;}
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