思路都在注释里，就是bfs层序遍历

java
class Solution {
    public List<Double> averageOfLevels(TreeNode root) {
        List<Double> averages = new ArrayList<>(); //答案数组 
        if (root == null) {  //特判
            return averages; 
        }

        Queue<TreeNode> queue = new LinkedList<>(); //bfs队列
        queue.offer(root); //根结点入队

        while (!queue.isEmpty()) {
              int levelSize = queue.size();
              double sum = 0;
              for (int i = 0; i < levelSize; i++) {  //入队并计算
                  TreeNode node = queue.poll();
                  sum += node.val;
                  if (node.left != null) {
                      queue.offer(node.left);
                  }
                  if (node.right != null) {
                      queue.offer(node.right);
                  }
              }
              averages.add(sum / levelSize);
          }

        return averages;

    }
}

java
public class Solution {
    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
                 int lenA = 0, lenB = 0;
         ListNode tempA = headA, tempB = headB;

         // 计算两个链表的长度
         while (tempA != null) {
              lenA++;
               tempA = tempA.next;
           }
         while (tempB != null) {
              lenB++;
               tempB = tempB.next;
           }

         // 对齐链表
         tempA = headA;
         tempB = headB;
         if (lenA > lenB) {
              for (int i = 0; i < lenA - lenB; i++) {
                   tempA = tempA.next;
               }
           } else {
               for (int i = 0; i < lenB - lenA; i++) {
                   tempB = tempB.next;
               }
           }

         // 遍历链表，找到相交节点
         while (tempA != null && tempB != null) {
               if (tempA == tempB) {
                   return tempA;
               }
               tempA = tempA.next;
               tempB = tempB.next;
           }

         return null;
       

    }
}

java
class Solution {
    public boolean isSymmetric(TreeNode root) {
        return isMirror(root.left,root.right);
    }

    private boolean isMirror(TreeNode t1, TreeNode t2) {
        if (t1 == null && t2 == null) return true;
        if (t1 == null || t2 == null) return false;
        return (t1.val == t2.val)
        && isMirror(t1.right, t2.left)
        && isMirror(t1.left, t2.right);
    }

}