Given a binary tree, return the level order traversal of its nodes’ values. (ie, from left to right, level by level).
For example: Given binary tree [3,9,20,null,null,15,7]
,
3
/ \
9 20
/ \
15 7
return its level order traversal as:
[
[3],
[9,20],
[15,7]
]
The code should be self-evident.
/**
* Definition for a binary tree node.
* function TreeNode(val) {
* this.val = val;
* this.left = this.right = null;
* }
*/
/**
* @param {TreeNode} root
* @return {number[][]}
*/
let levelOrder = function (root) {
if (!root) {
return [];
}
const result = [];
const queue = [NaN, root];
while (queue.length > 1) {
const node = queue.shift();
if (node !== node) {
result.push(queue.map((n) => n.val));
queue.push(NaN);
} else {
if (node.left) {
queue.push(node.left);
}
if (node.right) {
queue.push(node.right);
}
}
}
return result;
};
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☆: .。. o(≧▽≦)o .。.:☆☆: .。. o(≧▽≦)o .。.:☆