Leetcode-107(Java) 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,#,#,15,7}

,

3
/ \
9  20
/  \
15   7

return its bottom-up level order traversal as:

[
[15,7],
[9,20],
[3]
]

传送门：https://leetcode.com/problems/binary-tree-level-order-traversal-ii/
题目和第102题很像，只要把队列改成栈，逆序输出即可。

/**
* 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) {
//创建要返回的列表存放所有节点值
List<List<Integer>> list = new LinkedList<List<Integer>>();
//创建一个栈存放各层全部的节点值
if(root == null)
return list;
Stack<List<Integer>> slist = new Stack<List<Integer>>();
//创建一个栈存放一层的节点
Queue<TreeNode> currentLevel = new LinkedList<TreeNode>();
//先把根节点压入栈中
currentLevel.add(root);
while(!currentLevel.isEmpty())
{
int size = currentLevel.size();
//创建列表存放某一层的所有节点值
List<Integer> currentList = new LinkedList<Integer>();
for(int i = 0; i < size; i++)
{
TreeNode currentNode = currentLevel.poll();
currentList.add(currentNode.val);
if(currentNode.left != null)
currentLevel.add(currentNode.left);
if(currentNode.right != null)
currentLevel.add(currentNode.right);
}
slist.push(currentList);
}
while(!slist.isEmpty())
list.add(slist.pop());
return list;
}
}