第10题把二元查找树转换成排序的双向链表-优快云博客

题目：把二元查找树转换成排序的双向链表

要求：不能创建任何新的节点，只调整指针的指向

/ \

6 14

/ \ / \

4 8 12 16

转成双向链表

4=6=8=12=14=16

关于此题, 有一篇非常详细的文档，把这个问题讲述的非常清楚，英文好的朋友可以看一下这篇文章

http://download.youkuaiyun.com/detail/stevemarbo/4094319

#include<stdio.h> #include<stdlib.h> // binary search tree struct node { int data; struct node* small; struct node* large; }; void join(struct node* a, struct node* b) { a->large = b; b->small = a; } struct node* append(struct node* a, struct node* b) { struct node* aLast; struct node* bLast; if(a == NULL) return b; if(b == NULL) return a; aLast = a->small; bLast = b->small; join(aLast, b); join(bLast, a); return a; } struct node* treeToList(struct node* root) { struct node* aList; struct node* bList; if(root == NULL) return NULL; aList = treeToList(root->small); bList = treeToList(root->large); root->small = root; root->large = root; aList = append(aList, root); bList = append(aList, bList); return aList; } struct node* newNode(int data) { struct node* n = (struct node *)malloc(sizeof(struct node)); n->data = data; n->small = NULL; n->large = NULL; return n; } void treeInsert(struct node** rootRef, int data) { struct node* root = *rootRef; if(root == NULL) *rootRef = newNode(data); else { if(data <= root->data) treeInsert(&(root->small), data); else treeInsert(&(root->large), data); } } void printList(struct node* head) { struct node* current = head; while(current != NULL) { printf("%d ", current->data); current = current->large; if(current == head) break; } printf("\n"); } int main() { struct node* root = NULL; struct node* head; treeInsert(&root, 4); treeInsert(&root, 2); treeInsert(&root, 1); treeInsert(&root, 3); treeInsert(&root, 5); head = treeToList(root); printList(head); }