[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