LeetCode-in-All
102. Binary Tree Level Order Traversal
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: []
Constraints:
- The number of nodes in the tree is in the range
[0, 2000]. -1000 <= Node.val <= 1000
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 result = []
if (!root) {
return result
}
const queue = []
queue.push(root)
queue.push(null)
let level = []
while (queue.length > 0) {
root = queue.shift()
while (queue.length > 0 && root !== null) {
level.push(root.val)
if (root.left) {
queue.push(root.left)
}
if (root.right) {
queue.push(root.right)
}
root = queue.shift()
}
result.push(level)
level = []
if (queue.length > 0) {
queue.push(null)
}
}
return result
};
export { levelOrder }