1. 程式人生 > >求二叉樹中葉子結點的個數

求二叉樹中葉子結點的個數

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#include<iostream>#define N 63using namespace std;char str[] = "ab#d##c#e##";int i = -1;typedef
struct node{ struct node *leftChild; struct node *rightChild; char data;}BiTreeNode, *BiTree;//生成一個結點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); // 把結點連線成樹 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 = NULLelse {  T = new BiTreeNode;  T->data = c;  createBiTree2(T->leftChild);  createBiTree2(T->rightChild); }}int getLeafNode(BiTree T)if(NULL == T)  return 0if(NULL == T->leftChild && NULL == T->rightChild)  return 1return getLeafNode(T->leftChild) + getLeafNode(T->rightChild);}int main(){ BiTree T1; T1 = createBiTree1(); cout << getLeafNode(T1) << endl; BiTree T2; createBiTree2(T2); cout << getLeafNode(T2) << endlreturn 0;}


 

           

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