Journal Archive

/**
 * 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) {
    if (!root) return [];
    let result = [];
    let queue = [];
    // BFS approach, initial queue
    queue.push(root);
    while (queue.length !== 0) {
        let len = queue.length;
        let curLevel = [];
        for (let i = 0; i < len; i++) {
            let node = queue.shift();
            curLevel.push(node.val)
            if (node.left) {
                queue.push(node.left)
            }
            if (node.right) {
                queue.push(node.right)
            }
        }
        
        result.push(curLevel);
    }
    
    return result;
};

Day 27: Solving one of LeetCode problems

102. Binary Tree Level Order Traversal Difficulty - Medium

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: []