给定二叉树的前序和中序，判断是否可以构成一颗二叉树，如果可以输出后序

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很简单，注意边界条件

#include <stdio.h>const int kMaxN = 1000;int pre[kMaxN] = {0};int in[kMaxN] = {0};int post[kMaxN] = {0};int n = 0;bool Construct(int ps, int pe, int is, int ie, int s, int e) {  if (pe - ps != ie - is) return false;  int root = pre[ps];  post[e] = root;  int i;  for (i = is; i <= ie; i++) {    if (in[i] == root) break;  }  if (i > ie) {    return false;  }  int len1 = i - is;   int len2 = ie - i;  bool l = false;  bool r = false;  if (len1 == 0) {    l = true;  } else {    l = Construct(ps + 1, ps + len1, is, is + len1 - 1, s, s + len1 - 1);   }  if (len2 == 0) {    r = true;  } else {    r = Construct(ps + len1 + 1, pe, is + len1 + 1, ie, s + len1, e - 1);   }  return l && r;}int main() {  while (scanf("%d", &n) != EOF) {    for (int i = 0; i < n; i++) {      scanf("%d", &pre[i]);    }       for (int i = 0; i < n; i++) {      scanf("%d", &in[i]);    }       if (Construct(0, n - 1, 0, n - 1, 0, n - 1)) {      for (int i = 0; i < n; i++) {        printf("%d ", post[i]);      }      printf("\n");    } else {      printf("No\n");    }  }}