32 - II. 从上到下打印二叉树 II

comments: true
difficulty: 简单
edit_url: https://github.com/doocs/leetcode/edit/main/lcof/%E9%9D%A2%E8%AF%95%E9%A2%9832%20-%20II.%20%E4%BB%8E%E4%B8%8A%E5%88%B0%E4%B8%8B%E6%89%93%E5%8D%B0%E4%BA%8C%E5%8F%89%E6%A0%91%20II/README.md

面试题 32 - II. 从上到下打印二叉树 II

题目描述

从上到下按层打印二叉树，同一层的节点按从左到右的顺序打印，每一层打印到一行。

例如:
给定二叉树: [3,9,20,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7

返回其层次遍历结果：

[
  [3],
  [9,20],
  [15,7]
]

提示：

1. 节点总数 <= 1000

注意：本题与主站 102 题相同：https://leetcode.cn/problems/binary-tree-level-order-traversal/

解法

方法一：BFS

我们可以使用 BFS 的方法来解决这道题。首先将根节点入队，然后不断地进行以下操作，直到队列为空：

• 遍历当前队列中的所有节点，将它们的值存储到一个临时数组 $t$ 中，然后将它们的孩子节点入队。
• 将临时数组 $t$ 存储到答案数组中。

最后返回答案数组即可。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是二叉树的节点个数。

Python3

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None


class Solution:
    def levelOrder(self, root: TreeNode) -> List[List[int]]:
        ans = []
        if root is None:
            return ans
        q = deque([root])
        while q:
            t = [] #保存单层节点
            for _ in range(len(q)):
                node = q.popleft()
                t.append(node.val)
                if node.left:
                    q.append(node.left)
                if node.right:
                    q.append(node.right)
            ans.append(t)
        return ans

Java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> ans = new ArrayList<>();
        if (root == null) {
            return ans;
        }
        Deque<TreeNode> q = new ArrayDeque<>();
        q.offer(root);
        while (!q.isEmpty()) {
            List<Integer> t = new ArrayList<>();
            for (int n = q.size(); n > 0; --n) {
                TreeNode node = q.poll();
                t.add(node.val);
                if (node.left != null) {
                    q.offer(node.left);
                }
                if (node.right != null) {
                    q.offer(node.right);
                }
            }
            ans.add(t);
        }
        return ans;
    }
}

C++

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<vector<int>> levelOrder(TreeNode* root) {
        vector<vector<int>> ans;
        if (!root) return ans;
        queue<TreeNode*> q{{root}};
        while (!q.empty()) {
            vector<int> t;
            for (int n = q.size(); n; --n) {
                auto node = q.front();
                q.pop();
                t.push_back(node->val);
                if (node->left) q.push(node->left);
                if (node->right) q.push(node->right);
            }
            ans.push_back(t);
        }
        return ans;
    }
};

Go

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func levelOrder(root *TreeNode) (ans [][]int) {
	if root == nil {
		return
	}
	q := []*TreeNode{root}
	for len(q) > 0 {
		t := []int{}
		for n := len(q); n > 0; n-- {
			node := q[0]
			q = q[1:]
			t = append(t, node.Val)
			if node.Left != nil {
				q = append(q, node.Left)
			}
			if node.Right != nil {
				q = append(q, node.Right)
			}
		}
		ans = append(ans, t)
	}
	return
}

TypeScript

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function levelOrder(root: TreeNode | null): number[][] {
    const res = [];
    if (root == null) {
        return res;
    }
    const queue = [root];
    while (queue.length !== 0) {
        const n = queue.length;
        const tmp = new Array(n);
        for (let i = 0; i < n; i++) {
            const { val, left, right } = queue.shift();
            tmp[i] = val;
            left && queue.push(left);
            right && queue.push(right);
        }
        res.push(tmp);
    }
    return res;
}

Rust

// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::cell::RefCell;
use std::collections::VecDeque;
use std::rc::Rc;
impl Solution {
    pub fn level_order(root: Option<Rc<RefCell<TreeNode>>>) -> Vec<Vec<i32>> {
        let mut res = Vec::new();
        if root.is_none() {
            return res;
        }
        let mut queue = VecDeque::new();
        queue.push_back(root);
        while !queue.is_empty() {
            let n = queue.len();
            let mut vals = Vec::with_capacity(n);
            for _ in 0..n {
                let mut node = queue.pop_front().unwrap();
                let mut node = node.as_mut().unwrap().borrow_mut();
                vals.push(node.val);
                if node.left.is_some() {
                    queue.push_back(node.left.take());
                }
                if node.right.is_some() {
                    queue.push_back(node.right.take());
                }
            }
            res.push(vals);
        }
        res
    }
}

JavaScript

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number[][]}
 */
var levelOrder = function (root) {
    let ans = [];
    if (!root) {
        return ans;
    }
    let q = [root];
    while (q.length) {
        let t = [];
        for (let n = q.length; n; --n) {
            const { val, left, right } = q.shift();
            t.push(val);
            left && q.push(left);
            right && q.push(right);
        }
        ans.push(t);
    }
    return ans;
};

C#

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public IList<IList<int>> LevelOrder(TreeNode root) {
        if (root == null) {
            return new List<IList<int>>();
        }
        Queue<TreeNode> q = new Queue<TreeNode>();
        q.Enqueue(root);
        List<IList<int>> ans = new List<IList<int>>();
        while (q.Count != 0) {
            List<int> tmp = new List<int>();
            int x = q.Count;
            for (int i = 0; i < x; i++) {
                TreeNode node = q.Dequeue();
                tmp.Add(node.val);
                if (node.left != null) {
                    q.Enqueue(node.left);
                }
                if (node.right != null) {
                    q.Enqueue(node.right);
                }
            }
            ans.Add(tmp);
        }
        return ans;
    }
}