[leetcode]671. Second Minimum Node In a Binary Tree

[leetcode]671. Second Minimum Node In a Binary Tree

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Analysis

嘻嘻~—— [嘻嘻~]

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.
显然根节点是最小值，然后递归遍历其左右子树，第一个和根节点不相等的数就是第二大的数。

Implement

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int findSecondMinimumValue(TreeNode* root) {
        if(!root || !root->left)
            return -1;
        int val = root->val;
        int res = minValue(root, val);
        if(res != INT_MAX)
            return res;
        return -1;
    }
    int minValue(TreeNode* node, int val){
        if(node->val != val)
            return node->val;
        else if(node->val == val && node->left)
            return min(minValue(node->left, val), minValue(node->right, val));
        else
            return INT_MAX;
    }
};