Level Order Traversal of Binary Tree

class Solution {
public:
    vector<vector<int>> levelOrder(TreeNode* root) {
        vector<vector<int>> ans;
        if (!root) return ans;
        queue<TreeNode*> q;
        q.push(root);
        while (!q.empty()) {
            int sz = q.size();
            vector<int> level;
            level.reserve(sz);
            for (int i = 0; i < sz; ++i) {
                TreeNode* cur = q.front(); q.pop();
                level.push_back(cur->val);
                if (cur->left) q.push(cur->left);
                if (cur->right) q.push(cur->right);
            }
            ans.push_back(std::move(level));
        }
        return ans;
    }
};

func levelOrder(root *TreeNode) [][]int {
    ans := [][]int{}
    if root == nil {
        return ans
    }
    queue := []*TreeNode{root}
    for len(queue) > 0 {
        sz := len(queue)
        level := make([]int, 0, sz)
        for i := 0; i < sz; i++ {
            cur := queue[0]
            queue = queue[1:]
            level = append(level, cur.Val)
            if cur.Left != nil {
                queue = append(queue, cur.Left)
            }
            if cur.Right != nil {
                queue = append(queue, cur.Right)
            }
        }
        ans = append(ans, level)
    }
    return ans
}

from collections import deque
from typing import List, Optional

class Solution:
    def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
        ans: List[List[int]] = []
        if not root:
            return ans
        q = deque([root])
        while q:
            level = []
            for _ in range(len(q)):
                cur = q.popleft()
                level.append(cur.val)
                if cur.left:
                    q.append(cur.left)
                if cur.right:
                    q.append(cur.right)
            ans.append(level)
        return ans

var levelOrder = function(root) {
    const ans = [];
    if (!root) return ans;
    const q = [root];
    while (q.length > 0) {
        const sz = q.length;
        const level = [];
        for (let i = 0; i < sz; i++) {
            const cur = q.shift();
            level.push(cur.val);
            if (cur.left) q.push(cur.left);
            if (cur.right) q.push(cur.right);
        }
        ans.push(level);
    }
    return ans;
};

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()) {
            int sz = q.size();
            List<Integer> level = new ArrayList<>(sz);
            for (int i = 0; i < sz; i++) {
                TreeNode cur = q.poll();
                level.add(cur.val);
                if (cur.left != null) q.offer(cur.left);
                if (cur.right != null) q.offer(cur.right);
            }
            ans.add(level);
        }
        return ans;
    }
}

function levelOrder(root: TreeNode | null): number[][] {
    const ans: number[][] = [];
    if (!root) return ans;
    const q: TreeNode[] = [root];
    while (q.length > 0) {
        const sz = q.length;
        const level: number[] = [];
        for (let i = 0; i < sz; i++) {
            const cur = q.shift()!;
            level.push(cur.val);
            if (cur.left) q.push(cur.left);
            if (cur.right) q.push(cur.right);
        }
        ans.push(level);
    }
    return ans;
}