【LeetCode】222. Count Complete Tree Nodes

最新推荐文章于 2025-09-11 19:30:36 发布

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最新推荐文章于 2025-09-11 19:30:36 发布

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文章标签：数据结构与算法

原文链接：http://www.cnblogs.com/ganganloveu/p/4633843.html

本文介绍了一种高效计算完全二叉树中节点数量的方法。利用二叉树的特性，通过比较左右子树的深度，可以判断最后一层是否已满，从而避免遍历整棵树。如果最后一层满，则直接通过公式计算；否则采用递归计算。

Count Complete Tree Nodes

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.

完全二叉树相比普通二叉树，在计数时候一个trick在于，当最后一层是满的（最左深度等于最右深度），

就不需要通过遍历方式来计数，直接等于2^h - 1

/**
 * 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;
        TreeNode* l = root;
        int lefth = 0;
        while(l->left)
        {
            lefth ++;
            l = l->left;
        }
        TreeNode* r = root;
        int righth = 0;
        while(r->right)
        {
            righth ++;
            r = r->right;
        }
        if(lefth == righth)
            return pow(2.0, lefth+1) - 1;
        else
            return 1 + countNodes(root->left) + countNodes(root->right);
    }
};