102. Binary Tree Level Order Traversal

Binary Tree Level Order Traversal - LeetCodeLeetCode

DFS: better memory, faster

time: O(n)

time: O(h): average, O(n): worst

/**
 * 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[][]}
 */

let answer = [];

var levelOrder = function(root) {
    answer = [];

    findLevel(root);
    return answer;
};

function findLevel(node, level = 0) {
    if (node === null) return;

    answer[level] = answer[level] ?? [];
    answer[level].push(node.val);
    
    findLevel(node.left, level + 1);
    findLevel(node.right, level + 1);
}

iteration

time: O(n)

space: O(n)

/**
 * 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[][]}
 */

let answer = [];
let q = [];

var levelOrder = function(root) {
    answer = [];
    q = [[root, 0]];

    while (q.length > 0) {
        const [node, level] = q.shift();

        if (node === null) continue;

        if (answer[level] === undefined) {
            answer[level] = [];
        }
        
        answer[level].push(node.val);

        q.push([node.left, level + 1]);
        q.push([node.right, level + 1]);
    }

    return answer;
};