LeetCode 102. Binary Tree Level Order Traversal (javascript)

Given the root of a binary tree, return the level order traversal of its nodes’ values. (i.e., from left to right, level by level).

Example 1:

Input: root = [3,9,20,null,null,15,7]
Output: [[3],[9,20],[15,7]]

Example 2:

Input: root = [1]
Output: [[1]]

Example 3:

Input: root = []
Output: []

Constraints:

• The number of nodes in the tree is in the range [0, 2000].
• -1000 <= Node.val <= 1000

Idea:

Use BFS Template

Note: Don’t forget root === null case

Solution:

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number[][]}
 */
var levelOrder = function(root) {
    const res = [];
    if (root === null) return res;
    
    const queue = [];
    queue.push(root);
    while(queue.length !== 0) {
        let size = queue.length;
        let level = []; // store same level nodes
        while(size--) {
            let cur = queue.shift();
            level.push(cur.val);
            if (cur.left) queue.push(cur.left);
            if (cur.right) queue.push(cur.right);
        }
        res.push(level.concat());
    }
    return res;
};