当前位置: 首页 > 工具软件 > kint > 使用案例 >

2.6-17假设二叉树采用二叉链存储结构存放,结点值为 int 类型,设计一个递归算法求 二叉树 bt 中所有结点值大于等于 k 的结点个数

陶高扬
2023-12-01
#include "Btree.cpp"
#include <bits/stdc++.h>

int Nodenum(BTNode *bt,int k ) {
	if(bt == NULL)
		return 0;
	int lnum = Nodenum(bt->lchild,k);
	int rnum = Nodenum(bt->rchild,k);
	if(bt->data > k)
		return lnum + rnum + 1;
	else
		return lnum + rnum;
}

	int main() {
		BTNode *bt;
		int num;
		int a[]= {5,2,3,4,1,6};
		int b[]= {2,3,5,1,4,6};
		int n = sizeof(a) / sizeof(a[0]);
		bt = CreateBTree(a,b,n);
		DispBTree(bt);
		cout << endl;
		cout << "二叉树 bt 中所有结点值大于等于 3 的结点个数为:" << Nodenum(bt,3);
		DestroyBTree(bt);  // 释放结点 
		return 0;
	}
#include <bits/stdc++.h>

using namespace std;

typedef struct BNode {
	int data;
	struct BNode *lchild,*rchild;
}BTNode;

BTNode *CreateBTree(int a[],int b[], int n) {
	int k;
	if(n <= 0)
		return NULL; 
	BTNode *bt =(BTNode *) malloc (sizeof(BTNode));
	bt -> data = a[0];
	for(k = 0; k < n; k++)
		if(a[0] == b[k])
			break;
	//递归创建左子树
	bt -> lchild = CreateBTree(a+1, b, k);
	//递归创建左子树
	bt -> rchild = CreateBTree(a+k+1, b+k+1,n-1-k);
	return bt;
}

void DispBTree(BTNode *bt){
		if (bt != NULL){  //根左右 
			cout << bt->data << " ";
			DispBTree(bt->lchild);				
			DispBTree(bt->rchild);
		}
}	

void DestroyBTree(BTNode *bt){
	if(bt == NULL) {
		return ;
	}else{
		DestroyBTree(bt -> lchild);
		DestroyBTree(bt -> rchild);
		free(bt);
	}
}

 类似资料: