[Locked] Largest BST Subtree

Largest BST Subtree

Given a binary tree, find the largest subtree which is a Binary Search Tree (BST), where largest means subtree with largest number of nodes in it.

Note:
A subtree must include all of its descendants.
Here's an example:

    10
    / \
   5  15
  / \   \ 
 1   8   7

The Largest BST Subtree in this case is the highlighted one. 
The return value is the subtree's size, which is 3.

Follow up:
Can you figure out ways to solve it with O(n) time complexity?

分析：

　　典型树上的动态规划

代码：

//返回pair中4个值分别代表：是否是BST，BST的节点数，左边界，右边界
pair<pair<bool, int>, pair<int, int>> dfs(TreeNode *cur, int pval, int &maxl) {
    pair<int, int> initp(pval, pval);
    //为NULL，则返回真，两端值设为父节点的值便于下一步计算
    if(!cur)
        return make_pair(make_pair(true, 0), initp);
    //进行下一层遍历
    pair<pair<bool, int>, pair<int, int>> leftp, rightp;
    leftp = dfs(cur->left, cur->val, maxl);
    rightp = dfs(cur->right, cur->val, maxl);
    //判断是否为BST
    if(leftp.first.first && rightp.first.first && cur->val >= leftp.second.second && cur->val <= rightp.second.first) {
        int curlen = leftp.first.second + 1 + rightp.first.second;
        maxl = max(maxl, curlen);
        return make_pair(make_pair(true, curlen), make_pair(leftp.second.first, rightp.second.second));
    }
    return make_pair(make_pair(false, 0), initp);
}
int largestSubtree(TreeNode *root) {
    int maxl = INT_MIN;
    dfs(root, 0, maxl);
    return maxl;
}