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]
]

我的想法是廣度優先遍歷，把所有同層的找出來，值存起來放到結果數組中，再找下一層：

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number[][]}
 */
var levelOrder = function(root) {
    if (!root) {
        return [];
    }
    var result = [[root.val]];
    var seachingNodes = [root];
    var tempNodes,vals,num;
    while (seachingNodes.length!==0) {
        //console.log(seachingNodes);
        tempNodes = [];
        vals = [];
        num = seachingNodes.length;
        for (var i = 0; i < num; i++) {
            if (seachingNodes[i].left) {
                tempNodes.push(seachingNodes[i].left);
                vals.push(seachingNodes[i].left.val);
            }
            if (seachingNodes[i].right) {
                tempNodes.push(seachingNodes[i].right);
                vals.push(seachingNodes[i].right.val);
            }
        }
        if (vals.length!==0) {
            result[result.length]=(vals);
        }
        seachingNodes = tempNodes;
    }
    return result;
};