White Session - isomorphic -c language tree implementation .md

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First, understand the meaning of problems

Given two trees T1 and T2. If T1 can be interchanged by several times about the child becomes T2, the two trees we call a "homogeneous." Now given the two trees, you determine whether they are isomorphic.

Input format: Input information given binary 2:

• First given node in the tree of the tree in a row, followed by N rows

• I-th row corresponding to the i-th node number, the node gives letters stored in its left child node number, the right child node number

• If the child node is empty, then given in the respective position "-"

As shown below, there are various representation methods, we list the following two:

Second, the idea of ​​solving

1. Representation of Binary Tree
2. Build binary tree
3. Isomorphic discrimination

2.1 Representation of Binary Tree

Binary tree represents the array of structures: static list

/* c语言实现 */

#define MaxTree 10
#define ElementType char
#define Tree int
#define Null -1

struct TreeNode
{
  ElementType Element;
  Tree Left;
  Tree Right;
} T1[MaxTree], T2[MaxTree];

2.2 application framework to build

We need to design function:

• Read data to build a binary tree
• Binary isomorphic discrimination

/* c语言实现 */

int main():
{
  建二叉树1;
  建二叉树2;
  判别是否同构并输出;
  
  return 0;
}

int main()
{
  Tree R1, R2;
  
  R1 = BuildTree(T1);
  R2 = BuildTree(T2);
  if (Isomorphic(R1, R2)) printf("Yes\n");
  else printf("No\n");
  
  return 0;
}

2.3 How to build a binary tree

/* c语言实现 */

Tree BuildTree(struct TreeNode T[])
{
  ...;
  scanf("%d\n", &N); // 输入需要建立树的长度
  if (N) {
    ...;
    for (i=0; i<N; i++) {
      scanf("%c %c %c\n", &T[i].Element, &cl, &cr);
      ...;
    }
    ...;
    Root = ??? // 可以通过T[i]中没有任何结点的left(cl)和right(cr)指向他这个条件获取。
  }
  return Root;
}

/* c语言实现 */

Tree BuildTree(struct TreeNode T[])
{
  ...;
  scanf("%d\n", &N); // 输入需要建立树的长度
  if (N) {
    for (i=0; i<N; i++) check[i] = 0;
    for (i=0; i<N; i++) {
      scanf("%c %c %c\n", &T[i].Element, &cl, &cr);
      if (cl != '-'){
        T[i].Left = cl-'0';
        check[T[i].Left] = 1;
      }
      else T[i].Left = Null;
      ...;  // 对cr的对应处理
    }
    for (i=0; i<N; i++)
      if (!check[i]) break;
    Root = i; // 可以通过T[i]中没有任何结点的left(cl)和right(cr)指向他这个条件获取。
  }
  return Root;
}

2.4 How to determine two binary isomorphic

/* c语言实现 */

int Isomorphic(Tree R1, Tree R2)
{
  if ((R1 == Null) && (R2 == Null)) // 左右子树都为空
    return 1;
  if  (((R1==Null)&&(R2!=Null)) || ((R1!=Null)&&(R2==Null))) 
    return  0;  // 其中一颗子树为空
  if  (T1[R1].Element != T2[R2].Element) 
    return  0;  // 空结点为空
  if  ((T1[R1].Left == Null ) && ( T2[R2].Left == Null)) // 根的左右结点没有子树
    return  Isomorphic(T1[R1].Right, T2[R2].Right);
  if (((T1[R1].Left != Null) && (T2[R2].Left!=Null)) &&
      ((T1[T1[R1].Left].Element) == (T2[T2[R2].Left].Element))) // 左右子树不需要转换
  {
    return (Isomorphic(T1[R1].Left, T2[R2].Left) &&
            Isomorphic(T1[R1].Right, T2[R2].Right));
  }
  else { // 左右子树需要转换
    return (Isomorphic(T1[R1].Left, T2[R2].Right) &&
            Isomorphic(T1[R1].Right, T2[R2].Left));
  }
}