Binary Tree *Level* Order Traversal - 637. Average of Levels in Binary Tree

题目地址
典型的BFS了，只需要在遍历当前层级的时候记录下结点个数和值的累加和就好了。

    List result = new ArrayList<>();
   Queue q = new LinkedList<>();

• 注意new后面的具体数据类型：
  List = new ArrayList
  Queue = new LinkedList<>()

        int n = q.size();
        double sum = 0.0;
        for(int i = 0; i < n; i++) {

• q.size()是什么操作？
  记录每一层node数量.
  记住LinkedList的大小是用size()

            for (int i = 0; i < layerSize; i++ ) {
                TreeNode tempNode = q.poll();
                sum += tempNode.val;
                if (node.left != null) q.add(tempNode.left);
                if (node.right != null) q.add(tempNode.right);

• ArrayList, LinkedList - List添加元素都是add
  q.add(root);
  result.add(sum / levelSize)

• for loop 里初始化int又忘记了

            for (int i = 0; i < layerSize; i++ ) {

• node left, right 要先判断非null了以后再确定add

            for (int i = 0; i < layerSize; i++ ) {
                TreeNode tempNode = q.poll();
                sum += tempNode.val;
                if (node.left != null) q.add(tempNode.left);
                if (node.right != null) q.add(tempNode.right);

102. Binary Tree Level Order Traversal

忘记Base case了！

        if (root == null) return result;

class Solution {
    public List> levelOrder(TreeNode root) {
        List> result = new LinkedList<>();
        Queue q = new LinkedList<>();
        
        if (root == null) return result;
        
        q.offer(root);
        while(!q.isEmpty()){
            int n = q.size();
            List tempList = new ArrayList<>(); 
            for (int i = 0; i < n; i++) {
                TreeNode tempNode = q.poll();
                tempList.add(tempNode.val);
                if(tempNode.left != null) {q.offer(tempNode.left);}
                if(tempNode.right != null) {q.offer(tempNode.right);}
                
            }
            result.add(tempList);
        }
        return result;
    }

下一个目标是弄清楚：Top down和Bottom up

理解：
Bottom up - 是指不到最下面的叶节点，不会return（触底后反弹 - 下面得到的值一层层带回去）

Top down