二进制搜索树中最后N个节点的总和

Question

我想编写一个获得数字N和二叉搜索树的函数，然后该函数需要对树的最后N个节点的值求和。从较高到较低的节点值。我不能使用辅助函数或静态变量。

例如，如果函数获得了该二进制搜索树，并且N的值为3，则输出为：7 + 6 + 5。如果N为4，则为：7 + 6 + 5 + 3。

对算法有什么想法吗？

Answer 1

您可以简单地以相反的顺序访问树，这意味着从根开始并做三件事：

1. 访问右分支
2. 访问自身节点，并在需要时累加总和
3. 访问左分支

并在累积k个项目时停止迭代。

#include    <iostream>

struct Node {
    int value;
    Node* left = nullptr;
    Node* right = nullptr;
    Node(int v) : value(v) {}
};

// Visit the tree in reverse order and get sum of top k items.
int sumK(Node* root, int& k) {
    if (root == nullptr) return 0;
    int sum = 0;
    if (k > 0 && root->right) {
        sum += sumK(root->right, k);
    }
    if (k > 0) {
        sum += root->value;
        k--;
    }
    if (k > 0 && root->left) {
        sum += sumK(root->left, k);
    }
    return sum;
}

int main () {
    // The following code hard coded a tree matches the picture in your question.
    // I did not free the tree, so there will be memory leaking, but that is not relevant to this test.
    Node* root = new Node(5);
    root->right = new Node(6);
    root->right->right = new Node(7);
    root->left = new Node(3);
    root->left->right = new Node(4);
    root->left->left = new Node(2);
    root->left->left->left = new Node(1);
    // TODO: delete the tree after testing.

    int k = 1;
    std::cout << "k=" << k << " sum=" << sumK(root, k) << std::endl;
    k = 3;
    std::cout << "k=" << k << " sum=" << sumK(root, k) << std::endl;
    k = 4;
    std::cout << "k=" << k << " sum=" << sumK(root, k) << std::endl;
}

输出为：

k=1 sum=7
k=3 sum=18
k=4 sum=22