leetcode 671. Second Minimum Node In a Binary Tree

Given a non-empty special binary tree consisting of nodes with the non-negative value, where each node in this tree has exactly two or zero sub-node. If the node has two sub-nodes, then this node’s value is the smaller value among its two sub-nodes.

Given such a binary tree, you need to output the second minimum value in the set made of all the nodes’ value in the whole tree.

If no such second minimum value exists, output -1 instead.
分析题意可得思路，维护两个最小值min1和min2，然后用先序遍历，遍历整个二叉树，更新得到最后的结果，返回min2即得解。
最令人惊讶的是，这解法居然只要0ms…….

    int findSecondMinimumValue(TreeNode* root) {
        if (!root) return -1;
        int min1 = root->val;
        int min2 = INT_MAX;

        stack<TreeNode*> s;
        s.push(root);
        while (!s.empty()) {
            TreeNode* t = s.top();
            s.pop();
            if (t->val < min1) {
                min2 = min1;
                min1 = t->val;
            } else if (t->val != min1 && t->val < min2) {
                min2 = t->val;
            }

            if (t->left) s.push(t->left);
            if (t->right) s.push(t->right);
        }

        if (min2 == INT_MAX) return -1;

        return min2;
    }