| comments | difficulty | edit_url |
|---|---|---|
true |
简单 |
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：
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/
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
中，然后将它们的孩子节点入队。 - 将临时数组
存储到答案数组中。
最后返回答案数组即可。
时间复杂度
# 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 = []
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/**
* 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;
}
}/**
* 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;
}
};/**
* 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
}/**
* 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;
}// 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::rc::Rc;
use std::cell::RefCell;
use std::collections::VecDeque;
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
}
}/**
* 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;
};/**
* 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;
}
}