[LeetCode]Binary Tree Level Order Traversal II,解题报告

前言

中午容易犯困，做两道ACM有助于活跃脑细胞，

，汗了，csdn还可以添加qq标清，拜托支持一下markdown语法吧

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).

思路

题目不难，说是从底向上按层遍历二叉树，其实偷个懒，从上向下挨层遍历之后，然后对ArrayList里的元素来个逆序就OK了

从上向下挨层遍历二叉树，参考链接：java集合LinkeLlist模拟队列

ArrayList元素逆序，可以自己写个翻转的方法（双指针O(n)），也可以直接调用Collections类里的静态方法reverse

AC代码

/**
* Definition for binary tree
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public ArrayList<ArrayList<Integer>> levelOrderBottom(TreeNode root) {
ArrayList<ArrayList<Integer>> list = new ArrayList<ArrayList<Integer>>();
LinkedList<TreeNode> queue = new LinkedList<TreeNode>();
TreeNode p;

if (root == null) {
return list;
}

queue.addLast(root);

while (! queue.isEmpty()) {
ArrayList<Integer> sublist = new ArrayList<Integer>();
LinkedList<TreeNode> subqueue = new LinkedList<TreeNode>();

while (! queue.isEmpty()) {
p = queue.removeFirst();
sublist.add(p.val);

if (p.left != null) {
subqueue.addLast(p.left);
}

if (p.right != null) {
subqueue.addLast(p.right);
}
}

list.add(sublist);
queue.addAll(subqueue);
}

Collections.reverse(list);
return list;
}
}