[LeetCode] Count Complete Tree Nodes

最新推荐文章于 2024-09-12 01:58:26 发布

转载最新推荐文章于 2024-09-12 01:58:26 发布 · 140 阅读

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原文链接：https://yq.aliyun.com/articles/53058

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#数据结构与算法 #runtime

本文介绍了一种高效计算完全二叉树中节点数量的方法。通过比较左右子树的高度来判断是否为满二叉树，并据此计算节点总数。提供了一个C++实现示例。

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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.

解题思路

完全遍历会超时。可以从根节点开始分别计算左子树和右子树的高度，如果相等，则为满二叉树，结点个数为2n−1；如果高度不相等，则结点个数为左子树结点个数+右子树结点个数+1。（高度n是包含根节点的高度）

实现代码

// Runtime: 100 ms
/**
 * 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) {
        int leftHight = getLeft(root);
        int rightHeight = getRight(root);
        if (leftHight == rightHeight)
        {
            return pow(2, leftHight) - 1;
        }

        return 1 + countNodes(root->left) + countNodes(root->right);
    }

private:
    int getLeft(TreeNode *root)
    {
        int height = 0;
        while (root)
        {
            height++;
            root = root->left;
        }

        return height;
    }

    int getRight(TreeNode *root)
    {
        int height = 0;
        while (root)
        {
            height++;
            root = root->right;
        }

        return height;
    }
};