[LeetCode]102. Binary Tree Level Order Traversal

题目描述: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,null,null,15,7],
3
/ \
9 20
/ \
15 7
return its level order traversal as:
[
[3],
[9,20],
[15,7]
]

解题思路：
将树的节点遍历，并且记录每个节点所在的层数， 将层数相同的节点存放在同一个list中。list.add(level,new ArrayList<Integer>())

public List<List<Integer>> levelOrder(TreeNode root) {        List<List<Integer>> resList = new ArrayList<List<Integer>>();        if(root==null)return new ArrayList<List<Integer>>();        return levelOrder(resList,0,root);    }    public List<List<Integer>> levelOrder(List<List<Integer>> resList,int level,TreeNode root){        if(resList.size()<=level){            //增加一个子list            resList.add(level,new ArrayList<Integer>());        }        resList.get(level).add(root.val);        if(root.left!=null)levelOrder(resList,level+1,root.left);        if(root.right!=null)levelOrder(resList,level+1,root.right);        return resList;    }