(三)二元查找樹轉換雙向鏈表

題意：不引入新節點，只允許指針操作，實現二元查找樹轉換爲相應雙向鏈表。

    10

   /    /

  6    14

 / /    /  /

4  8 12 16

轉換成雙向鏈表

4=6=8=10=12=14=16

 

 

分析：

方法0：轉換後的雙向鏈表正好爲原樹的中序遍歷，所以引入一個新指針遞歸遍歷即可。

方法1：從10節點看，它的前趨、後繼分別是左子樹的最右節點與右子樹的最左節點，所以找到前趨與後繼保存，遞歸實現即可。

 

 

代碼中驗證Btree建立成功分別用先序、中序與後序來遍歷。然後又用到了Btree拷貝方法驗證兩種tree2list方法。

C++實現：

#include <iostream> #include <stack> using namespace std; struct btree{ btree(int v):val(v), l(NULL), r(NULL){} int val; btree *l, *r; }; void preorder(btree* head){ btree* tmp = head; stack<btree*> s; while(!s.empty() || tmp!=NULL){ while(tmp!=NULL){ s.push(tmp); cout << s.top()->val << "/t"; tmp=tmp->l; } tmp=s.top()->r; s.pop(); } } void inorder(btree* head){ btree* tmp = head; stack<btree*> s; while(!s.empty() || tmp!=NULL){ while(tmp!=NULL){ s.push(tmp); tmp=tmp->l; } tmp=s.top()->r; cout << s.top()->val << "/t"; s.pop(); } } void postorder(btree* head){ btree* tmp = head; stack<btree*> s; stack<bool> f; while(!s.empty() || tmp!=NULL){ while(tmp!=NULL){ s.push(tmp); f.push(false); tmp=tmp->l; } if(f.top()==false){ tmp=s.top()->r; f.pop(); f.push(true); }else{ cout << s.top()->val << "/t"; s.pop(); f.pop(); } } } void walklist(btree* head){ btree* tmp=head; while(tmp!=NULL){ cout << tmp->val << "/t"; tmp=tmp->r; } } //way0: inorder travel void tree2list(btree* bt,btree*& list) { if(bt->l!=NULL) tree2list(bt->l, list); if(list!=NULL){ list->r = bt; bt->l = list; list = bt; }else{ list=bt; } if(bt->r) tree2list(bt->r, list); } //way1: find the pre and post btree* getRight(btree* bt){ if(bt==NULL) return bt; while(bt->r!=NULL){ bt = bt->r; } return bt; } btree* getLeft(btree* bt){ if(bt==NULL) return bt; while(bt->l!=NULL){ bt = bt->l; } return bt; } void tree2list1(btree*& bt){ btree *pre, *post; if(bt==NULL) return; pre = getRight(bt->l); post = getLeft(bt->r); if(bt->l!=NULL) tree2list1(bt->l); if(bt->r!=NULL) tree2list1(bt->r); bt->l = pre; if(pre!=NULL) pre->r = bt; bt->r = post; if(post!=NULL) post->l = bt; } btree* treeCopy(btree* bt){ if(bt==NULL) return bt; btree* tmp = new btree(bt->val); if(bt->l!=NULL) tmp->l = treeCopy(bt->l); if(bt->r!=NULL) tmp->r = treeCopy(bt->r); return tmp; } int main() { /*creat tree*/ btree *n1 = new btree(10); btree *n2 = new btree(6); btree *n3 = new btree(14); btree *n4 = new btree(4); btree *n5 = new btree(8); btree *n6 = new btree(12); btree *n7 = new btree(16); //creat Btree n1->l = n2, n1->r = n3; n2->l = n4, n2->r = n5; n3->l = n6, n3->r = n7; btree *n8 = treeCopy(n1); //travel Btree cout << "Preorder:/t"; preorder(n1); cout << endl; cout << "Midorder:/t"; inorder(n1); cout << endl; cout << "Postorder:/t"; postorder(n1); cout << endl; //Way0: Btree to list btree* list=NULL; tree2list(n1,list); list = n1; while(list!=NULL && list->l!=NULL) list=list->l; cout << "DoubleList:/t"; walklist(list); cout << endl; //Way1: Btree to list tree2list1(n8); list = n8; while(list!=NULL && list->l!=NULL) list=list->l; cout << "DoubleList1:/t"; walklist(list); cout << endl; return 0; }   

 

 

運行結果：