Leet Code OJ 102. Binary Tree Level Order Traversal [Difficulty: Easy]

题目:
Given a binary tree, return the level order traversal of its nodes’ values. (ie, from left to right, level by level).

For example:
Given binary tree {3,9,20,#,#,15,7},

return its level order traversal as:
[
[3],
[9,20],
[15,7]
]

翻译：
给定一个二叉树，返回它的节点值的层序遍历（即从左到右，一层一层的）。

分析：
采用递归的方式，先返回根节点，然后递归调用左右节点，把左右节点的结果进行按照题目要求的形式拼装。另外注意一些递归过程中节点为空的情况。

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>> 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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