Rebuilding a binary tree

Enter the result of the forward and middle traversal of a binary tree. Create a New Binary Tree. Assume that the input results do not contain repeated numbers. For example, input pre-order traversal sequences {1, 2, 4, 7, 3, 5, 6, 8} and Middle-order traversal sequences {4, 7, 2, 1, 5, 3, 8, 6 }, then, the binary tree is rebuilt and its post-sequential traversal sequence is output. Input: the input may contain multiple test examples. For each test case, the first input behavior is an integer n (1 <= n <= 1000), indicating the number of nodes in the binary tree. The second line of the input contains n integers (the range of each element a is (1 <= a <= 1000), which indicates the pre-order traversal sequence of the binary tree. The third line of the input contains n integers (where the range of each element a is (1 <= a <= 1000): indicates the ordinal traversal sequence of the binary tree. Output: corresponds to each test case, and outputs a line: if the pre-order and Middle-order traversal sequences given in the question can constitute a binary tree, n integers are output, represents the post-order traversal sequence of a binary tree. Each element is followed by spaces. If the pre-order and mid-order traversal sequences given in the question cannot constitute a binary tree, "No" is output ". Example input: 81 2 4 7 3 5 6 84 7 2 1 5 3 8 681 2 4 7 5 6 84 1 2 5 3 8 6 sample output: 7 4 2 5 8 6 3 1 No code AC: Thought: Use sub-governance to build a tree! [Cpp] # include <stdio. h> # include <stdlib. h> typedef struct tree {int id; struct tree * lc; struct tree * rc;} tree, * p_tree; p_tree root; int * pre, * mid; int creat_tree (int low, int high, p_tree * t, int m_low, int m_high) {int I, count = 0, f1, f2; int flag; if (high-low! = M_high-m_low) {return 0;} if (low> high) // detail {return 1;} (* t) = (p_tree) malloc (sizeof (tree )); (* t)-> id = pre [low]; (* t)-> lc = NULL; (* t)-> rc = NULL; // if (low = high) // details // {// return 1; //} flag = 0; for (I = m_low; I <= m_high; I ++) {if (mid [I] = pre [low]) {flag = 1; break;} else {count ++ ;}} if (flag) {f1 = creat_tree (low + 1, low + count, & (* t)-> lc), m_low, M_low + count-1); if (! F1) {return 0;} f2 = creat_tree (low + count + 1, high, & (* t)-> rc), m_low + count + 1, m_high ); return f2;} else {return 0;} void out_put (p_tree t) {if (t) {out_put (t-> lc); out_put (t-> rc ); printf ("% d", t-> id) ;}} int main () {int I, n; while (scanf ("% d", & n )! = EOF) {pre = (int *) malloc (sizeof (int) * n); mid = (int *) malloc (sizeof (int) * n ); for (I = 0; I <n; I ++) {scanf ("% d", & pre [I]) ;}for (I = 0; I <n; I ++) {scanf ("% d", & mid [I]);} www.2cto.com if (creat_tree (0, n-1, & root, 0, n-1) {out_put (root);} else {printf ("No");} printf ("\ n");} return 0 ;}