如何在不使用动态内存分配的情况下创建基于指针的二叉树？

#include <iostream>

const int MAX_NODES = 100; // 最大节点数

struct Node {
    int data;
    int left;
    int right;
};

Node tree[MAX_NODES]; // 静态数组作为二叉树的存储结构
int root = -1; // 根节点的索引

int createNode(int data) {
    static int index = 0; // 静态变量用于记录节点索引
    tree[index].data = data;
    tree[index].left = -1;
    tree[index].right = -1;
    return index++;
}

void insertNode(int data) {
    if (root == -1) {
        root = createNode(data);
        return;
    }

    int currentNode = root;
    while (true) {
        if (data < tree[currentNode].data) {
            if (tree[currentNode].left == -1) {
                tree[currentNode].left = createNode(data);
                break;
            } else {
                currentNode = tree[currentNode].left;
            }
        } else {
            if (tree[currentNode].right == -1) {
                tree[currentNode].right = createNode(data);
                break;
            } else {
                currentNode = tree[currentNode].right;
            }
        }
    }
}

void inorderTraversal(int currentNode) {
    if (currentNode == -1) {
        return;
    }

    inorderTraversal(tree[currentNode].left);
    std::cout << tree[currentNode].data << " ";
    inorderTraversal(tree[currentNode].right);
}

int main() {
    insertNode(5);
    insertNode(3);
    insertNode(7);
    insertNode(1);
    insertNode(4);

    std::cout << "Inorder Traversal: ";
    inorderTraversal(root);
    std::cout << std::endl;

    return 0;
}