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;
}
}