二叉树与森林的转换



    关键思路：利用递归建立节点完成转换。首先我们观察二叉树的结构，包含第一个孩子，及其兄弟。关键在于：那么我们在构造递归时，每层recursion中都必须找到对应sibling和firstchild。对于双亲表示法，我们很难找到孩子，对于孩子表示法，我们很难找到双亲（因为找到双亲是找到本节点兄弟的唯一方法，或在传参时也传入父节点指针）。

   不同形式间转换的技巧：对于要被转换的结构，需要传入针对它单次recursion完成结构内部所有成员构造对应的转换的结构的信息。就能完成一层递归。同时应注意递归结束的条件!

A) 孩子表示法生成二叉树算法:

typedef struct BiTree{ElemType data;struct BiTree *firstchild, *sibling;}BTnode, *BTree;typedef struct Lnode{int child;struct Lnode *next;}Lnode, *CList;typedef struct Cbox{ElemType data;CList firstchild;}Cbox, *Cnode;typedef struct CTnode{Cnode node;int n, r;}CTree;void TransformChildTreeToBinaryTree(CTree CT, BTree &BT, int preroot, int root){if (root == -1){BT = NULL; return;}BT = (BTree)malloc(sizeof(BTnode));BT->data = CT.node[root].data;int rsave;if (CT.node[root].firstchild){rsave = CT.node[root].firstchild->child;}else{rsave = -1;}TransformChildTreeToBinaryTree(CT, BT->firstchild, root, rsave);if (preroot == -1){BT->sibling = NULL; return;}CList p = CT.node[preroot].firstchild;while (p && p->data != root){p = p->next;}if (p->next){rsave = p->next->child;}else{rsave = -1;}TransformChildTreeToBinaryTree(CT, BT->sibling, preroot, rsave);return;}

B) 孩子双亲表示法生成二叉树(类似于孩子表示法):

typedef struct BTnode{ElemType data;struct BTnode *firstchild, *sibling;}BTnode, *BTree;typedef struct Cnode{int child;struct Cnode *next;}Cnode, *CList;typedef struct CTnode{int parent;ElemType data;CList firstchild;}CTnode, *CTbox;typedef struct PCTnode{CTbox node;int r, n;}PCTree;void TransformPCTreeToBiTree(PCTree PC, BTree &BT, int root){if (root == -1){BT = NULL; return;}BT = (BTree)malloc(sizeof(BTnode));BT->data = PC.node[root].data;int rsave, tmp;if (PC.node[root].firstchild){rsave = PC.node[root].firstchide->child;}else{rsave = -1;}TransformPCTreeToBiTree(PC, BT->firstchild, rsave);tmp = PC.node[root].parent;if (tmp == -1){BT->sibling = NULL; return;}else{LList p = PC.node[tmp].firstchild;while (p && p->child != root){p = p->next;}if (p->next){rsave = p->next->child;}else{rsave = -1;}TransformPCTreeToBiTree(PC, BT->sibling, rsave);return;}}

C)同构链式孩子表示法生成二叉树:

typedef struct BTnode{ElemType data;struct BTnode *firstchild, *sibling;}BTnode, *BTree;#define DEGREE 5typedef struct LCTnode{ElemType data;struct LCTnode *child[DEGREE];}LCTnode, *LCTree;BTree InitialTransLCTree2BiTree(LCTree root){if (!root){return NULL;}BTree BTroot = (BTree)malloc(sizeof(BTnode));BTroot->data = LCTree->data;BTroot->sibling = NULL;TransformLinkedChildTreeToBinaryTree(BTroot->firstchild, root, 0);return BTroot;}void TransformLinkedChildTreeToBinaryTree(BTree &BT, LCTree LCT, int i){if (i >= DEGREE || !LCT->child[i]){BT = NULL; return;}BT = (BTree)malloc(sizeof(BTnode));BT->data = LCT->child[i]->data;TransformLinkedChildTreeToBinaryTree(BT, LCT->sibling, i + 1);TransformLinkedChildTreeToBinaryTree(BT->child[i], LCT->firstchild, 0);return;}

D) 异构链式孩子表示法生成二叉树:

typedef struct BTnode{ElemType data;struct BTnode *firstchild, *sibling;}BTnode, *BTree;#ifdef DEGREE#undef DEGREE#define DEGREE 5#endiftypedef struct LCTnode{ElemType data;struct LCTnode * *child;int degree;}LCTnode, *LCTree;void TransformLCTree2BinaryTree(LCTree LCT, BTree &BT, int i){if (i >= LCT->degree || !LCT->child[i]){BT = NULL; return;}BT = (BTree)malloc(sizeof(BTnode));BT->data = LCT->child[i]->data;TransformLCTree2BinaryTree(LCT, BT->sibling, i + 1);TransformLCTree2BinaryTree(LCT->child[i], BT->firstchild, 0);return;}//Another Method:void TransformLCTree2BinaryTree2(LCTree LCT, BTree &BT){if (!LCT->child[0]){BT = NULL; return;}BTree p, q;for (int i = 0; i < LCT->degree && LCT->child[i]; ++i){BTree q = (BTree)malloc(sizeof(BTnode));q->data = LCT->child[i]->data;q->sibling = NULL;q->firstchild = NULL;if (!BT){BT = q;p = q;}else{q->sibling = p->sibling; p->sibling = q;p = q;}TransformLCTree2BinaryTree2(LCT->child[i], q->firstchild);return;}}

D中共有两种方法的思路不同:前者是按单个节点完成递归, 而后者则是按照层来递归(虽然二者都是深度优先), 后者更像是图的遍历方法.

上面的算法仅供参考, 如果大家有疑惑或者更好的算法欢迎讨论,如有错误请大家指出和批评!