力扣每日一题——完全二叉树的节点个数

题目连接

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int countLevel(TreeNode* root){//返回二叉树的高度
        int level=0;
        while(root){//节点不空时
            level++;
            root=root->left;//不断由左子树向下伸展
        }
        return level;
    }
    int countNodes(TreeNode* root) {
        if(!root){//空节点
            return 0;
        }
        int left = countLevel(root->left);
        int right = countLevel(root->right);
        if(left==right){//左子树一定是满二叉树
            return countNodes(root->right)+(1<<left);
        }
        else{//left=right+1右子树是满二叉树
            return countNodes(root->left)+(1<<right);
        }
    }
};

参考题解链接