[LeetCode] 107. Binary Tree Level Order Traversal II Java

题目：

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]
]

 题意及分析：给出一颗二叉树广度遍历的结果，从叶节点到根节点。和前面根节点到叶节点类似，只是结果用Collections.reverse反转一下即可。

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>> levelOrderBottom(TreeNode root) {
        List<List<Integer>> res = new ArrayList<>();
        if(root==null) return res;
        List<Integer> list = new ArrayList<>();
        Queue<TreeNode> queue=new LinkedList<>();
        queue.offer(root);
        Queue<TreeNode> queue1=new LinkedList<>();
        while(!queue.isEmpty()||!queue1.isEmpty()){
            queue1.clear();
            while(!queue.isEmpty()){        //遍历一层
                TreeNode now = queue.poll();
                list.add(now.val);
                if(now.left!=null)
                    queue1.offer(now.left);
                if(now.right!=null)
                    queue1.offer(now.right);
            }
            queue=new LinkedList<>(queue1);       //进行下一层的遍历
            res.add(new ArrayList<>(list));
            list.clear();       //清空
        }
        Collections.reverse(res);
        return res;
    }
}