leetcode: Binary Tree Level Order Traversal II

问题描述：

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]
]

原问题链接：https://leetcode.com/problems/binary-tree-level-order-traversal-ii/

 

问题分析

　　这个问题和前面树的层次化遍历的问题基本上一样，唯一有一点差别的地方就是在遍历完之后将结果倒置一下就可以了。

　　详细的代码实现如下：

 

/**
 * 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>> result = new ArrayList<>();
        if(root == null) return result;
        List<TreeNode> list = new ArrayList<>();
        list.add(root);
        while(!list.isEmpty()) {
            List<Integer> intList = new ArrayList<>();
            List<TreeNode> temp = new ArrayList<>();
            for(TreeNode node : list) {
                intList.add(node.val);
                if(node.left != null) temp.add(node.left);
                if(node.right != null) temp.add(node.right);
            }
            result.add(intList);
            list = temp;
        }
        Collections.reverse(result);
        return result;
    }
}