完全二叉树节点个数

1、没利用完全二叉树性质的递归

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    Queue<TreeNode> q = new LinkedList<>();
    public int countNodes(TreeNode root) {
    if(root == null) return 0;    
    return countNodes(root.left) + countNodes(root.right) + 1;
        
    }
}

2、因为完全二叉树只有最后一层不是满的。
1.1、左子树不是满二叉树，右子树自然就是满二叉树了

1.2、左子树是满二叉树，右子树不一定。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int countNodes(TreeNode root) {
        if(root == null){
           return 0;
        } 
        int left = countLevel(root.left);
        int right = countLevel(root.right);
        if(left == right){//左子树是满二叉树
            return countNodes(root.right) + (1<<left);//左子树加上根节点数目刚好是2^left,用位运算快一点
        }else{
            return countNodes(root.left) + (1<<right);//同理
        }
    }
    private int countLevel(TreeNode root){//可以帮助判断是否左子树是满二叉树
        int level = 0;
        while(root != null){
            level++;
            root = root.left;
        }
        return level;
    }
}