Created
August 9, 2021 02:14
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Binary Tree Level Order Traversal LeetCode
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/* | |
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 | |
*/ | |
/** | |
* 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 []; | |
const arr = [root], result = []; | |
while (arr.length) { | |
const size = arr.length, values = []; | |
for (let i = 0; i < size; i++) { | |
const node = arr.shift() | |
values.push(node.val) | |
if (node.left) arr.push(node.left) | |
if (node.right) arr.push(node.right) | |
} | |
result.push(values) | |
} | |
return result | |
}; | |
const root = [3,9,20,null,null,15,7]; | |
console.log(levelOrder(root)); |
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