个人记录-LeetCode 107. Binary Tree Level Order Traversal II

问题：
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]]

这个问题要求从叶子节点往根节点输出。
其是直接广度优先遍历，依次输出每一层节点的值，
每次将下一层节点的值插入到上一层的前面即可。

代码示例：

/** * 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) {        //保存最后的结果        List<List<Integer>> rst = new LinkedList<>();        if(root == null) {            return rst;        }        //保存节点        Queue<TreeNode> queue = new LinkedList<>();        queue.offer(root);        //保存每一层节点的值        List<Integer> subList;        while(!queue.isEmpty()){            //得到当前层节点的数量            int num = queue.size();            subList = new LinkedList<>();            //将下一层的节点加入到queue中            //同时将当前层节点的值加入到subList            for(int i = 0; i < num; i++) {                if(queue.peek().left != null) {                    queue.offer(queue.peek().left);                }                if(queue.peek().right != null) {                    queue.offer(queue.peek().right);                }                subList.add(queue.poll().val);            }            //subList插入到index=0的位置，即最前面            rst.add(0, subList);        }        return rst;    }}