王道数据结构应用题强化表3.1.1-3.1.6

3.1.1 

度为0结点数 = 度为2的结点+1

二叉树的第 i 层上至多有 2^(i-1) 个结点 (i≥1)

深度为 k 的二叉树至多有 2^k - 1 个结点 (k≥1)

3.1.2

#include"stdio.h" 
#include <iostream>
using namespace std;

#define MAXSIZE 100

typedef struct arrayTree
{
    //array从下标一开始，下标0标记为长度
    int array[MAXSIZE+1];
}ArrayTree;

ArrayTree a;

// 找到父节点
int findFather (int i)
{
    if (i == 1) return -1;
    return i/2;
}

// 找到左节点
int leftChild (int i)
{
    if(i*2<=MAXSIZE)
        return  i*2;
    return -1;
}

// 找到右节点
int rightChild (int i)
{
    if(i*2+1<=MAXSIZE)
    {
        return i*2+1;
    }
    return -1;
}

// 3.1.6 先序遍历
void preOrder(int i) {
    if (i > MAXSIZE || a.array[i] == 0) return;
    
    cout << a.array[i] << " ";  // 访问根节点
    int left = leftChild(i);
    if (left != -1) preOrder(left);  // 遍历左子树
    int right = rightChild(i);
    if (right != -1) preOrder(right); // 遍历右子树
}

// 3.1.6 中序遍历
void inOrder(int i) {
    if (i > MAXSIZE || a.array[i] == 0) return;
    
    int left = leftChild(i);
    if (left != -1) inOrder(left);  // 遍历左子树
    cout << a.array[i] << " ";   // 访问根节点
    int right = rightChild(i);
    if (right != -1) inOrder(right); // 遍历右子树
}

// 3.1.6 后序遍历
void postOrder(int i) {
    if (i > MAXSIZE || a.array[i] == 0) return;
    
    int left = leftChild(i);
    if (left != -1) postOrder(left);  // 遍历左子树
    int right = rightChild(i);
    if (right != -1) postOrder(right); // 遍历右子树
    cout << a.array[i] << " ";     // 访问根节点
}


int main() 
{ 
    a.array[0] = 7;  // 共7个节点
    a.array[1] = 1;
    a.array[2] = 2;
    a.array[3] = 3;
    a.array[4] = 4;
    a.array[5] = 5;
    a.array[6] = 6;
    a.array[7] = 7;
    
    // 测试父节点查找
    printf("节点5的父节点编号: %d\n", findFather(5));
    printf("节点3的父节点编号: %d\n", findFather(3));
    
    // 测试遍历
    printf("先序遍历: ");
    preOrder(1);
    printf("\n");
    
    printf("中序遍历: ");
    inOrder(1);
    printf("\n");
    
    printf("后序遍历: ");
    postOrder(1);
    printf("\n");
    
 
}