[LeetCode] BFS - 102. Binary Tree Level Order Traversal

문제

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
•  

코드

/**
 * 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 queue = [];
    const result = [];

    if(root !== null) {
        queue.push([root, 0])
    }

    while(queue.length) {
        const [node, level] = queue.shift();
        if(!result[level]) result[level] = [];
        if(node && node.val !== null)result[level].push(node.val);
        if(node && node.left) queue.push([node.left, level + 1])
        if(node && node.right) queue.push([node.right, level + 1])
    }

    return result
};

// 혹은

// If root is null return an empty array
if(!root) return []

const queue = [root] // initialize the queue with root
const levels = [] // declare output array

while(queue.length !== 0){
   const queueLength = queue.length // Get the length prior to dequeueing
   const currLevel = [] // Declare this level
   // loop through to exhaust all options and only to include nodes at currLevel
   for(let i = 0; i < queueLength; i++){
       // Get next node
       const current = queue.shift()

       if(current.left){
           queue.push(current.left)
       }
       if(current.right){
           queue.push(current.right)
       }
       // After we add left and right for current, we add to currLevel
       currLevel.push(current.val)
   }
   // Level has been finished. Push into output array
   levels.push(currLevel)
}
return levels
}