首页 > 代码库 > 107. Binary Tree Level Order Traversal II

107. Binary Tree Level Order Traversal II

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,null,null,15,7],

    3   /   9  20    /     15   7

 

return its bottom-up level order traversal as:

[  [15,7],  [9,20],  [3]]
思路:跟I一样,把结果插到arraylist但开头就可以。
/** * 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>> res= new ArrayList<List<Integer>>();        if(root==null)        {            return res;        }                Queue<TreeNode> check=new LinkedList<TreeNode>();                check.offer(root);        while(!check.isEmpty())        {            List<Integer> curl=new ArrayList<Integer>();            for(int i=check.size()-1;i>=0;i--)            {            TreeNode find=check.poll();            curl.add(find.val);            if(find.left!=null)            {                check.offer(find.left);            }            if(find.right!=null)            {                check.offer(find.right);            }            }            res.add(0,curl);        }        return res;    }    }

 

107. Binary Tree Level Order Traversal II