102. Binary Tree Level Order Traversal

和之前一题很像，把之前一题的解法答案reverse即可

Java

/**
 * 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 LinkedList<List<Integer>>();
        count(result,root,0);
        Collections.reverse(result);
        return result;
    }
    
     public void count(List<List<Integer>> list,TreeNode root,int level)
    {
        if(root==null) return;
        if(level>=list.size())
        {
            list.add(0,new LinkedList<Integer>());
        }
        count(list,root.left,level+1);
        count(list,root.right,level+1);
        list.get(list.size()-level-1).add(root.val);
    }
}

Java，貌似不需要reverse，递归的时候list中加值的顺序可以改

public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> res = new ArrayList<List<Integer>>();
        levelHelper(res, root, 0);
        return res;
    }
    
    public void levelHelper(List<List<Integer>> res, TreeNode root, int height) {
        if (root == null) return;
        if (height >= res.size()) {
            res.add(new LinkedList<Integer>());
        }
        res.get(height).add(root.val);
        levelHelper(res, root.left, height+1);
        levelHelper(res, root.right, height+1);
    }