给你一棵 完全二叉树 的根节点 root ，求出该树的节点个数。

完全二叉树 的定义如下：在完全二叉树中，除了最底层节点可能没填满外，其余每层节点数都达到最大值，并且最下面一层的节点都集中在该层最左边的若干位置。若最底层为第 h 层，则该层包含 1~ 2h 个节点。

示例 1：

输入：root = [1,2,3,4,5,6]
输出：6

示例 2：

输入：root = []
输出：0

示例 3：

输入：root = [1]
输出：1

方法一 

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
 //普通二叉树计算节点数量，后续遍历
class Solution {
    public int countNodes(TreeNode root) {
        if(root == null) return 0;
        int leftcount = countNodes(root.left);
        int rightcount = countNodes(root.right);
        return leftcount+rightcount+1;
    }
}

方法二 

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public int countNodes(TreeNode root) {
        if(root==null) return 0;
        TreeNode leftnode = root.left;
        TreeNode rightnode = root.right;
        int leftnum = 0;
        int rightnum = 0;
        //判断是否为满二叉树，可以减少遍历节点的次数
        while(leftnode!=null){
            leftnum++;
            leftnode = leftnode.left;
        }
        while(rightnode!=null){
            rightnum++;
            rightnode = rightnode.right;
        }
        if(rightnum==leftnum){
            return (2<<leftnum )-1;//满二叉树2^n-1;
        }
        return countNodes(root.left)+countNodes(root.right)+1;

    }
}