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 100typedef 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");}