leecode_222 Count Complete Tree Nodes

最新推荐文章于 2024-10-28 00:00:00 发布

qq_31552435

最新推荐文章于 2024-10-28 00:00:00 发布

阅读量208

点赞数

CC 4.0 BY-SA版权

分类专栏： LeeCode

本文链接：https://blog.youkuaiyun.com/qq_31552435/article/details/51601048

LeeCode 专栏收录该内容

35 篇文章

订阅专栏

本文介绍了一种高效计算完全二叉树节点数量的方法，通过比较左右子树的高度判断是否为完全二叉树，并利用数学公式计算节点数，避免了传统递归的高时间复杂度。

Given a complete binary tree, count the number of nodes.

Definition of a complete binary tree from Wikipedia:

In a complete binary tree every level, except possibly the last, is completely filled, and all nodes in the last level are as far left as possible. It can have between 1 and 2h nodes inclusive at the last level h.

用暴力法, recursive求会超时 O(N). 如果从某节点一直向左的高度 = 一直向右的高度, 那么以该节点为root的子树一定是complete binary tree. 而 complete binary tree的节点数,可以用公式算出 2^h - 1. 如果高度不相等, 则递归调用 return countNode(left) + countNode(right) + 1. 复杂度为O(h^2)

c++ 代码：

/**
 * 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 countNodes(TreeNode* root) {
        if(root==NULL) return 0;  
          
        int l = getLeft(root) + 1;  
        int r = getRight(root) + 1;  
          
        if(l==r) {  
            return (2<<(l-1)) - 1;  
        } else {  
            return countNodes(root->left) + countNodes(root->right) + 1;  
        }  
    }
    
    int getLeft(TreeNode* root) {  
        int count = 0;  
        while(root->left!=NULL) {  
            root = root->left;  
            ++count;  
        }  
        return count;  
    }  
    
    int getRight(TreeNode* root) {  
        int count = 0;  
        while(root->right!=NULL) {  
            root = root->right;  
            ++count;  
        }  
        return count;  
    }  
};