[LeetCode] 107. Binary Tree Level Order Traversal II
Given the root of a binary tree, return the bottom-up level order traversal of its nodes' values. (i.e., from left to right, level by level from leaf to root).
Example 1:
Input: root = [3,9,20,null,null,15,7]
Output: [[15,7],[9,20],[3]]
Example 2:
Input: root = [1]
Output: [[1]]
Example 3:
Input: root = []
Output: []
Constraints:
The number of nodes in the tree is in the range [0, 2000].
-1000 <= Node.val <= 1000
二叉树的层序遍历 II。
给你二叉树的根节点 root ,返回其节点值 自底向上的层序遍历 。 (即按从叶子节点所在层到根节点所在的层,逐层从左向右遍历)
思路
这道题跟102题唯一的区别是每一层的节点整理好以后,是加到结果集的头部而不是尾部。这里我只提供 BFS 实现。
复杂度
时间O(n)
空间O(n^2)
代码
Java实现
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode() {} * TreeNode(int val) { this.val = val; } * TreeNode(int val, TreeNode left, TreeNode right) { * this.val = val; * this.left = left; * this.right = right; * } * } */ class Solution { public List<List<Integer>> levelOrderBottom(TreeNode root) { List<List<Integer>> res = new ArrayList<>(); // corner case if (root == null) { return res; } // normal case Queue<TreeNode> queue = new LinkedList<>(); queue.offer(root); while (!queue.isEmpty()) { int size = queue.size(); List<Integer> list = new ArrayList<>(); for (int i = 0; i < size; i++) { TreeNode cur = queue.poll(); list.add(cur.val); if (cur.left != null) { queue.offer(cur.left); } if (cur.right != null) { queue.offer(cur.right); } } res.add(0, list); } return res; } }
相关题目
102. Binary Tree Level Order Traversal 107. Binary Tree Level Order Traversal II 429. N-ary Tree Level Order Traversal
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