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;
    }
}