二叉树基本操作实现(二叉树的链式存储)
1 #include <iostream> 2 using namespace std; 3 typedef struct bintree_node 4 { 5 int data; 6 bintree_node *lchild; 7 bintree_node *rchild; 8 }bintree_node; 9 10 typedef struct stack 11 { 12 bintree_node *data[100]; 13 int tag[100];//为栈中每个元素设置标记,用于后序遍历 14 int top; 15 }seqstack; 16 17 void push(seqstack *s, bintree_node *p) 18 { 19 s->data[++s->top] = p; 20 } 21 22 bintree_node *pop(seqstack *s) 23 { 24 if(s->top != -1) 25 { 26 s->top--; 27 return (s->data[s->top + 1]); 28 } 29 else 30 return NULL; 31 } 32 33 bintree_node *bintree_create() 34 { 35 bintree_node *p = (bintree_node *)malloc(sizeof(bintree_node)); 36 if(NULL == p) 37 { 38 cout << "alloc memory error!!!" << endl; 39 return NULL; 40 } 41 cin >> (p->data); 42 if(p->data == -1) 43 { 44 return NULL; 45 } 46 p->lchild = bintree_create(); 47 p->rchild = bintree_create(); 48 return p; 49 } 50 51 void prebintree(bintree_node *p)//前序遍历递归实现 52 { 53 if(p) 54 { 55 cout << p->data << "\t"; 56 prebintree(p->lchild); 57 prebintree(p->rchild); 58 } 59 } 60 void prebintree1(bintree_node *p)//前序遍历非递归实现 61 { 62 seqstack s; 63 s.top = -1; 64 while ((p) || (s.top != -1)) 65 { 66 while (p) 67 { 68 cout << p->data << "\t"; 69 s.top++; 70 s.data[s.top] = p; 71 p = p->lchild; 72 } 73 if(s.top > -1) 74 { 75 p = pop(&s); 76 p = p->rchild; 77 } 78 } 79 } 80 void levelTraverse(bintree_node *p) 81 { 82 bintree_node *queue[20]; 83 bintree_node *tmp; 84 int front = 0, rear = 0;//表示队头指针和队尾指针 85 if(p) 86 { 87 tmp = p;//根指针入队 88 queue[rear] = tmp; 89 rear = (rear + 1)%20;//队尾指针加一对20取余,可实现循环队列,合理利用空间 90 while(front != rear)//队不空 91 { 92 tmp = queue[front];//出队,将值赋给tmp 93 cout << tmp->data; 94 front = (front + 1)%20; 95 if(tmp -> lchild)//如果tmp有左子树,将左子树入队 96 { 97 queue[rear] = tmp->lchild; 98 rear = (rear + 1)%20; 99 } 100 if(tmp->rchild)//如果tmp有右子树,将右子树入队 101 { 102 queue[rear] = tmp->rchild; 103 rear = (rear + 1)%20; 104 } 105 } 106 } 107 } 108 int main() 109 { 110 bintree_node *p = bintree_create(); 111 prebintree1(p); 112 levelTraverse(p); 113 cout << endl; 114 return 0; 115 }