Uva--548 （二叉树，遍历，重建）
2014-06-19 00:27:02
题意&思路：给出一棵二叉树的中序遍历和后序遍历，让你求出从根节点到任意叶节点路径上节点值总合的最小值。思路就是根据两个遍历建树，DFS搜索之。
非常感谢http://blog.csdn.net/acvcla/article/details/27554837?reload的精简代码，借鉴了！
1 #include <cstdio> 2 #include <cstring> 3 #include <iostream> 4 using namespace std; 5 const int maxn = 100000005; 6 7 int in[10005],post[10005],tmin,tag; 8 struct node{ 9 int val; 10 node *left,*right; 11 node(){ 12 val = 0; 13 left = right = NULL; 14 } 15 }; 16 node * Build_tree(int *in,int *post,int cnt){ 17 if(cnt <= 0) 18 return NULL; 19 node *next = new node; 20 int pos; 21 for(pos = 0; pos < cnt ; ++pos) 22 if(in[pos] == post[0]) 23 break; 24 if(pos > 0) 25 next->left = Build_tree(in,post + cnt - pos,pos); 26 if(pos < cnt - 1) 27 next->right = Build_tree(in + pos + 1,post + 1,cnt - pos - 1); 28 next->val = post[0]; 29 30 return next; 31 } 32 void Dfs(node *tem,int sum){ 33 if(tem == NULL) 34 return; 35 if(tem->left == NULL && tem->right == NULL){ 36 if(sum + tem->val < tmin){ 37 tmin = sum + tem->val; 38 tag = tem->val; 39 } 40 } 41 Dfs(tem->left,sum + tem->val); 42 Dfs(tem->right,sum + tem->val); 43 } 44 void Delete(node *tem){ 45 if(!tem) 46 return; 47 if(tem->left != NULL) 48 Delete(tem->left); 49 if(tem->right != NULL) 50 Delete(tem->right); 51 delete tem; 52 tem = NULL; //Important 53 } 54 int main(){ 55 char t; 56 node *root; 57 while(scanf("%d",&in[0]) == 1){ 58 root = NULL; 59 int cnt = 1; 60 while((t = getchar()) != '\n'){ 61 scanf("%d",in + cnt); 62 ++cnt; 63 } 64 for(int i = cnt - 1; i >= 0; --i){ 65 scanf("%d",post + i); 66 } 67 tmin = tag = maxn; 68 tag = 0; 69 root = Build_tree(in,post,cnt); 70 Dfs(root,0); 71 printf("%d\n",tag); 72 73 Delete(root); 74 } 75 return 0; 76 }
