1385重建二叉树

剑指offer 

题目1385：重建二叉树

print1前序遍历，判断二叉树是否合法，即与输入的二叉树是否匹配，若不匹配输出No；

print2后序遍历输出答案；

 1 #include <stdio.h>
 2 int idx[1001],b[1001];
 3 typedef  struct node
 4 {
 5     struct node * lson,*rson;
 6     int value;
 7 }node,*root;
 8 node tree[1001];
 9 root build(int l,int r)
10 {
11     int i,midx,mval=10000;
12     if(l > r)return NULL;
13     for(i=l;i<=r;i++)if(mval>b[i]){mval = b[i],midx = i;}
14     tree[midx].lson = build(l, midx-1);
15     tree[midx].rson = build(midx+1,r);
16     tree[midx].value = mval;
17     return (&tree[midx]);
18     
19 }
20 int num;
21 int print1(root ROOT)
22 {
23     if(ROOT == NULL)return 0;
24     //printf("%d ",ROOT->value);
25     if(num++!=idx[ROOT->value])return -1;
26     int f1 = print1(ROOT->lson);
27     int f2 = print1(ROOT->rson);
28     return (f1<f2?f1:f2);
29     
30 }
31 void print2(root ROOT)
32 {
33     if(ROOT == NULL)return;
34     print2(ROOT->lson);
35     print2(ROOT->rson);
36     printf("%d ",ROOT->value);
37     
38 }
39 
40 int main(int argc, const char * argv[])
41 {
42     int n,i,a;
43     while(scanf("%d",&n)!=EOF)
44     {
45         for(i=0;i<n;i++)
46         {
47             scanf("%d",&a);
48             idx[a] = i;
49         }
50         for(i=0;i<n;i++)
51             scanf("%d",b+i);
52         root ROOT = build(0, n-1);
53         num =0;
54         if(print1(ROOT)==-1)printf("No\n");
55         else
56         {
57             print2(ROOT);
58             printf("\n");
59         }  
60     }
61 }

 

posted on 2013-07-01 21:49  4.5.6  阅读(377)  评论(0)  编辑  收藏  举报