LeetCode--No.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]]

代码我直接照着Binary Tree Level Order Traversal I 改的
就是将最后的list倒过来。。。

/** * 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) {        LinkedList<List<Integer>> tmp = new LinkedList<List<Integer>>();        List<List<Integer>> res = new ArrayList<List<Integer>>();                if(root == null)            return res;        List<TreeNode> currlayer= new ArrayList<TreeNode>();        currlayer.add(root);        while(!currlayer.isEmpty()){            List<TreeNode> nextlayer = new ArrayList<TreeNode>();            List<Integer> currvalue = new ArrayList<Integer>();            for(TreeNode node : currlayer){                currvalue.add(node.val);                if(node.left != null)                    nextlayer.add(node.left);                if(node.right != null)                    nextlayer.add(node.right);            }            tmp.add(currvalue);            currlayer = nextlayer;        }        while(!tmp.isEmpty()){            List<Integer> list = tmp.removeLast();            res.add(list);        }        return res;    }}