1、沒利用完全二叉樹性質的遞迴

/**
 * 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、因為完全二叉樹只有最後一層不是滿的。
1.1、左子樹不是滿二叉樹，右子樹自然就是滿二叉樹了

1.2、左子樹是滿二叉樹，右子樹不一定。

/**
 * 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){//左子樹是滿二叉樹
            return countNodes(root.right) + (1<<left);//左子樹加上根節點數目剛好是2^left,用位運算快一點
        }else{
            return countNodes(root.left) + (1<<right);//同理
        }
    }
    private int countLevel(TreeNode root){//可以幫助判斷是否左子樹是滿二叉樹
        int level = 0;
        while(root != null){
            level++;
            root = root.left;
        }
        return level;
    }
}