comments: true
difficulty: 中等
edit_url: https://github.com/doocs/leetcode/edit/main/lcof/面试题32 - I. 从上到下打印二叉树/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返回：
[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 ) O(n) O(n)，空间复杂度 O ( n ) O(n) O(n)。其中 n n n 为二叉树的节点数。
Python3
python
# 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[int]:
ans = []
if root is None:
return ans
q = 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 ansJava
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++
cpp
/**
* 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
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
ts
/**
* 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
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
js
/**
* 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#
cs
/**
* 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
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
}
}