剑指offer(12)-二叉搜索树的后序遍历序列[数据结构]九度1367


http://www.acmerblog.com/offer12-2969.html
输入一个整数数组,判断该数组是不是某二叉搜索树的后序遍历的结果。如果是则输出Yes,否则输出No。假设输入的数组的任意两个数字都互不相同。
对应二叉树的问题普遍是用递归解决。二叉搜索树也就是二分查找树。左子树的每个节点都比根节点小,右子树的每个节点都比根节点大。

以 5 7 6 9 11 10 8 为例: 根节点为8,左子树为 5 7 6  又子树为 11 10。
所以,任务就是先找到左子树。

程序中是这样找的:
int k = start; //k为右子树的开始位置(左子树都是比根节点小的数)
while(arr[k] < arr[end] && k <= end-1) k++;
此时,arr[k] == 11。即右子树开始的位置。 接下来判断右子树是否是符合要求的,即都比根节点大。
09int n, arr[10001], j;
10bool check(int start, int end){
11    //任意3个或3个以下的数都符合
12    if(start >= end-2) return true;
13    int k = start; //k为右子树的开始位置(左子树都是比根节点小的数)
14    while(arr[k] < arr[end] && k <= end-1) k++;
15    j = k;
16    //只要右子树中有比根节点小的就不符合
17    while(j < end) if(arr[j++] < arr[end]) return false;
18    return check(start,k-1) && check(k,end-1); //分别判断左右子树
19}
20int main() {
21    while(scanf("%d", &n) != EOF){
22        for(int i=0; i<n; i++) scanf("%d", &arr[i]);
23        if( check(0,n-1) ) puts("Yes");
24        else puts("No");
25    }
26    return 0;
27}
http://bylijinnan.iteye.com/blog/1300758

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