Enter Two Binary Trees, A and B, and determine whether B is a sub-structure. The nine-degree questions are described as follows:

Description: Enter Two Binary Trees A and B to determine whether B is a sub-structure. Input: the input may contain multiple test examples. The input ends with EOF. For each test case, the input first line is an integer n, m (1 <= n <= 1000, 1 <= m <= ): N indicates the number of nodes of the Binary Tree A to be input (the number of nodes starts from 1), and M indicates the number of nodes of the Binary Tree B to be input (the number of nodes starts from 1 ). The next row contains N numbers, each representing the I-th element of the tree, and the next n rows. The first Ki number represents the number of children on the I-th node, next there are ki trees, representing the node I child node number. The next m + 1 line is the same as the description of tree. Output: corresponds to each test case. If B is a sub-tree of a, "yes" (excluding quotation marks) is output ). Otherwise, "no" is output (excluding quotation marks ). Example input: 7 38 8 7 9 2 4 72 2 32 4 5002 6 7008 9 22 2 300

My Preface provides specific ideas in "offoffoffer", but the binary tree data structure definition provided by him obviously cannot meet the Nine-degree problem, because I really cannot come up with other methods besides using struct arrays to represent trees. Note that I am talking about pure C implementation, c ++ is different from other languages. Here, my binary tree node is defined as follows:

// Binary Tree node defines struct btree {int value; int lchild, rchild ;};

The structure of tree A and Tree B is as follows to determine whether Tree B is a sub-tree of tree A: to find whether tree A has a sub-tree similar to tree B, there are two steps:

- Locate the node R with the same value as the root node of Tree B in tree.
- Then, determine whether the sub-tree with R as the root node in tree A contains a sub-structure such as tree B.

It sounds like a recursive process (pure C). Now I only use pure C, PHP, and shell programming languages. I can basically understand C ++ and Java, we still need to get involved in a wider range. First, we need to lay a solid foundation for algorithms and computer systems, and then start the first step. In Tree A, we need to find the same value as that of the B root node, in fact, is the pre-order traversal of the tree, it is recommended to recursive, convenient (PS: Non recursion is nothing more than a stack storage node, no technical content, refer to the link: http://blog.csdn.net/zinss26914/article/details/8450952)

/*** Step 1: traverse tree A to find out if there is a subtree equal to the B root node */INT judgechildtree (struct btree * ahead, int NUMA, struct btree * bhead, int numb) {int flag = 0; If (NUMA! =-1 & numb! =-1) {If (ahead [NUMA]. value = bhead [numb]. Value) Flag = doestree1hastree2 (ahead, NUMA, bhead, numb); If (! Flag & ahead [NUMA]. lchild! =-1) Flag = judgechildtree (ahead, ahead [NUMA]. lchild, bhead, numb); If (! Flag & ahead [NUMA]. rchild! =-1) Flag = judgechildtree (ahead, ahead [NUMA]. rchild, bhead, numb);} return flag ;}

Step 2: Determine whether the subtree with R as the root node of tree A has the same node as that of Tree B.

/*** Step 2: judge whether tree A has B sub-structure */INT doestree1hastree2 (struct btree * ahead, int NUMA, struct btree * bhead, int numb) {If (numb =-1) return 1; if (NUMA =-1) return 0; If (ahead [NUMA]. value! = Bhead [numb]. value) return 0; Return (doestree1hastree2 (ahead, ahead [NUMA]. lchild, bhead, bhead [numb]. lchild) & doestree1hastree2 (ahead, ahead [NUMA]. rchild, bhead, bhead [numb]. rchild ));}

Complete code (9 degree system problems, I can't determine whether my code is AC, I think it is OK)

# Include <stdio. h> # include <stdlib. h> // define the struct btree {int value; int lchild, rchild ;}; // The maximum number of knots of tree A and Tree B, int n, m; /*** Step 2: judge whether tree A has B sub-structure */INT doestree1hastree2 (struct btree * ahead, int NUMA, struct btree * bhead, int numb) {If (numb =-1) return 1; if (NUMA =-1) return 0; If (ahead [NUMA]. value! = Bhead [numb]. value) return 0; Return (doestree1hastree2 (ahead, ahead [NUMA]. lchild, bhead, bhead [numb]. lchild) & doestree1hastree2 (ahead, ahead [NUMA]. rchild, bhead, bhead [numb]. rchild);}/*** Step 1: traverse tree A to check whether there is a subtree equal to the B root node */INT judgechildtree (struct btree * ahead, int NUMA, struct btree * bhead, int numb) {int flag = 0; If (NUMA! =-1 & numb! =-1) {If (ahead [NUMA]. value = bhead [numb]. Value) Flag = doestree1hastree2 (ahead, NUMA, bhead, numb); If (! Flag & ahead [NUMA]. lchild! =-1) Flag = judgechildtree (ahead, ahead [NUMA]. lchild, bhead, numb); If (! Flag & ahead [NUMA]. rchild! =-1) Flag = judgechildtree (ahead, ahead [NUMA]. rchild, bhead, numb);} return flag;} int main (void) {int I, Data, Count, left, right, flag; struct btree * ahead, * bhead; while (scanf ("% d", & N, & M )! = EOF) {// get the node value of tree a ahead = (struct btree *) malloc (sizeof (struct btree) * n); for (I = 0; I <N; I ++) {scanf ("% d", & data); ahead [I]. value = data; ahead [I]. lchild = ahead [I]. rchild =-1 ;}for (I = 0; I <n; I ++) {scanf ("% d", & COUNT); If (COUNT = 0) {continue;} else {If (COUNT = 1) {scanf ("% d", & left); ahead [I]. lchild = left-1;} else {scanf ("% d", & left, & right); ahead [I]. lchild = left-1; ahead [I]. rchild = right-1 ;}}// obtain the node value bhead of Tree B = (struct btree *) malloc (sizeof (struct btree) * m ); for (I = 0; I <m; I ++) {scanf ("% d", & data); bhead [I]. value = data; bhead [I]. lchild = bhead [I]. rchild =-1 ;}for (I = 0; I <m; I ++) {scanf ("% d", & COUNT); If (COUNT = 0) {continue;} else {If (COUNT = 1) {scanf ("% d", & left); bhead [I]. lchild = left-1;} else {scanf ("% d", & left, & right); bhead [I]. lchild = left-1; bhead [I]. rchild = right-1 ;}}// determine whether the subtree of Tree B is a if (n = 0 | M = 0) {printf ("NO \ n"); continue;} flag = judgechildtree (ahead, 0, bhead, 0); If (FLAG) printf ("Yes \ n "); elseprintf ("NO \ n"); free (ahead); free (bhead);} return 0 ;}