32 - I. 从上到下打印二叉树

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difficulty: 中等
edit_url: https://github.com/doocs/leetcode/edit/main/lcof/%E9%9D%A2%E8%AF%95%E9%A2%9832%20-%20I.%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/README.md

面试题 32 - I. 从上到下打印二叉树

题目描述

从上到下打印出二叉树的每个节点，同一层的节点按照从左到右的顺序打印。

例如:
给定二叉树: [3,9,20,null,null,15,7],

    3/ \9  20/  \15   7

[3,9,20,15,7]

提示：

节点总数 <= 1000

解法

方法一：BFS

我们可以通过 BFS 遍历二叉树，将每一层的节点值存入数组中，最后返回数组即可。

时间复杂度 $O (n)$ ，空间复杂度 $O (n)$ 。其中 $n$ 为二叉树的节点数。

Python3

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = Noneclass Solution:def levelOrder(self, root: TreeNode) -> List[int]:ans = []if root is None:return ansq = deque([root])while q:#保证层序下去for _ in range(len(q)):node = q.popleft()ans.append(node.val)if node.left:q.append(node.left)if node.right:q.append(node.right)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 int[] levelOrder(TreeNode root) {if (root == null) {return new int[] {};}Deque<TreeNode> q = new ArrayDeque<>();q.offer(root);List<Integer> res = new ArrayList<>();while (!q.isEmpty()) {for (int n = q.size(); n > 0; --n) {TreeNode node = q.poll();res.add(node.val);if (node.left != null) {q.offer(node.left);}if (node.right != null) {q.offer(node.right);}}}int[] ans = new int[res.size()];for (int i = 0; i < ans.length; ++i) {ans[i] = res.get(i);}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<int> levelOrder(TreeNode* root) {if (!root) {return {};}vector<int> ans;queue<TreeNode*> q{{root}};while (!q.empty()) {for (int n = q.size(); n; --n) {auto node = q.front();q.pop();ans.push_back(node->val);if (node->left) {q.push(node->left);}if (node->right) {q.push(node->right);}}}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 {for n := len(q); n > 0; n-- {node := q[0]q = q[1:]ans = append(ans, node.Val)if node.Left != nil {q = append(q, node.Left)}if node.Right != nil {q = append(q, node.Right)}}}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 ans: number[] = [];if (!root) {return ans;}const q: TreeNode[] = [root];while (q.length) {const t: TreeNode[] = [];for (const { val, left, right } of q) {ans.push(val);left && t.push(left);right && t.push(right);}q.splice(0, q.length, ...t);}return ans;
}

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<i32> {let mut ans = Vec::new();let mut q = VecDeque::new();if let Some(node) = root {q.push_back(node);}while let Some(node) = q.pop_front() {let mut node = node.borrow_mut();ans.push(node.val);if let Some(l) = node.left.take() {q.push_back(l);}if let Some(r) = node.right.take() {q.push_back(r);}}ans}
}

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) {const ans = [];if (!root) {return ans;}const q = [root];while (q.length) {const t = [];for (const { val, left, right } of q) {ans.push(val);left && t.push(left);right && t.push(right);}q.splice(0, q.length, ...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 int[] LevelOrder(TreeNode root) {if (root == null) {return new int[]{};}Queue<TreeNode> q = new Queue<TreeNode>();q.Enqueue(root);List<int> ans = new List<int>();while (q.Count != 0) {int x = q.Count;for (int i = 0; i < x; i++) {TreeNode node = q.Dequeue();ans.Add(node.val);if (node.left != null) {q.Enqueue(node.left);}if (node.right != null) {q.Enqueue(node.right);}}}return ans.ToArray();}
}

Swift

/* public class TreeNode {
*     var val: Int
*     var left: TreeNode?
*     var right: TreeNode?
*     init(_ val: Int) {
*         self.val = val
*         self.left = nil
*         self.right = nil
*     }
* }
*/class Solution {func levelOrder(_ root: TreeNode?) -> [Int] {guard let root = root else {return []}var queue: [TreeNode] = [root]var result: [Int] = []while !queue.isEmpty {let node = queue.removeFirst()result.append(node.val)if let left = node.left {queue.append(left)}if let right = node.right {queue.append(right)}}return result}
}