求二叉树中单分支结点的个数

本文主要是介绍求二叉树中单分支结点的个数，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

#include<iostream>
#define N 15using namespace std;char str[] = "ab#d##c#e##";
int i = -1;typedef struct node
{struct node *leftChild;struct node *rightChild;char data;
}BiTreeNode, *BiTree;//生成一个结点
BiTreeNode *createNode(int i)
{BiTreeNode * q = new BiTreeNode;q->leftChild = NULL;q->rightChild = NULL;q->data = i;return q;
}BiTree createBiTree1()
{BiTreeNode *p[N] = {NULL};int i;for(i = 0; i < N; i++)p[i] = createNode(i + 1);// 把结点连接成树for(i = 0; i < N/2; i++){p[i]->leftChild = p[i * 2 + 1];p[i]->rightChild = p[i * 2 + 2];}return p[0];
}void createBiTree2(BiTree &T)
{i++;char c;if(str[i] && '#' == (c = str[i]))T = NULL;else{T = new BiTreeNode;T->data = c;createBiTree2(T->leftChild);createBiTree2(T->rightChild);}
}int getSingleBranchNode(BiTree T)
{if(NULL == T)return 0;if(NULL == T->leftChild && NULL == T->rightChild)return 0;if(NULL != T->leftChild && NULL != T->rightChild)return getSingleBranchNode(T->leftChild) + getSingleBranchNode(T->rightChild);return 1 + getSingleBranchNode(T->leftChild) + getSingleBranchNode(T->rightChild);}int main()
{BiTree T1;T1 = createBiTree1();cout << getSingleBranchNode(T1) << endl;BiTree T2;createBiTree2(T2);cout << getSingleBranchNode(T2) << endl;return 0;
}

 

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