Day53 二叉树的层序遍历

news2025/5/11 12:32:20

给你二叉树的根节点 root ，返回其节点值的层序遍历。（即逐层地，从左到右访问所有节点）。

在这里插入图片描述

/**
 * 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>> levelOrder(TreeNode root) {
        List<List<Integer>> result = new ArrayList<>();
        if (root == null) return result;
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            int levelSize = queue.size();
            List<Integer> currentLevel = new ArrayList<>();
            for (int i = 0; i < levelSize; i++) {
                TreeNode node = queue.poll();
                currentLevel.add(node.val);
                if (node.left != null) queue.offer(node.left);
                if (node.right != null) queue.offer(node.right);
            }
            result.add(currentLevel);
        }
        return result;
    }
}

时间复杂度：O(n)
空间复杂度：O(n)（最坏情况下队列存储所有叶子节点）

public List<List<Integer>> levelOrder(TreeNode root) {
    List<List<Integer>> result = new ArrayList<>();
    traverse(root, 0, result);
    return result;
}

private void traverse(TreeNode node, int level, List<List<Integer>> result) {
    if (node == null) return;
    
    if (result.size() == level) {
        result.add(new ArrayList<>());
    }
    
    result.get(level).add(node.val);
    
    traverse(node.left, level + 1, result);
    traverse(node.right, level + 1, result);
}