671. Second Minimum Node In a Binary Tree

题目描述

Given a non-empty special binary tree consisting of nodes with the non-negative value, where each node in this tree has exactlytwo 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.

Example 1:

Input:     2   / \  2   5     / \    5   7Output: 5Explanation: The smallest value is 2, the second smallest value is 5.

Example 2:

Input:     2   / \  2   2Output: -1Explanation: The smallest value is 2, but there isn't any second smallest value.

解题思路：

这个题目首先遍历一遍求出最大值，然后将min与min_next赋值给最大值。然后二次遍历，找到次小值就可以了。

时间复杂度是O（n）

解答代码：

/**
 * 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 min,minn=-1;
    
    void get_minn(TreeNode* root)
    {
        if(root)
        {
            if(minn < root->val)
                min = minn = root->val;
            get_minn(root->left);
            get_minn(root->right);
        }
    }
    
    void preorder(TreeNode* root)
    {
        if(root)
        {
            if(root->val < min )
            {
                minn = min;
                min = root->val;
            }
            if(root->val <minn && root->val >min)
                minn = root->val;
            
            preorder(root->left);
            preorder(root->right);
        }
    }
    int findSecondMinimumValue(TreeNode* root) {
        if( root==NULL || (root->left==NULL&&root->right==NULL&&root!=NULL) )
            return -1;
        int min = root->val;
        
        get_minn(root);
        
        preorder(root);
        
        if(minn == min)
            return -1;        
        
        return minn;
    }
    
};