树的直径

编程之美3.8。这题让我想起了另外一道著名的题目：lowest common root. 两题都是典型的树之上的DP。下面为实现：

struct Node {
    int v;
    Node *left, *right;
};

size_t diameter(Node* root, size_t& depth) {
    depth = 0;
    if (!root) return 0;
    size_t ldepth, rdepth;
    size_t ldiameter = diameter(root->left, ldepth);
    size_t rdiameter = diameter(root->right, rdepth);
    depth = max(ldepth, rdepth) + 1;
    return max(ldepth + rdepth + 2, max(ldiameter, rdiameter));
}

size_t lcr(Node* root, Node* a, Node* b, Node*& v) {
    if (!root) return 0;
    size_t nl = lcr(root->left, a, b, v);
    if (nl == 2) return 2;
    size_t nr = lcr(root->right, a, b, v);
    if (nr == 2) return 2;
    size_t nm = nl + nr + (root == a || root == b);
    if (nm == 2) v = root;
    return nm;
}