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. More formally, the property root.val = min(root.left.val, root.right.val) always holds.

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   7

Output: 5
Explanation: The smallest value is 2, the second smallest value is 5.

Example 2:

Input: 
    2
   / 
  2   2

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

Accepted

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Submissions

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因为整个树有可能值全部一样,因此最坏情况要全部遍历才能知道是否有第二大的值. 

另外一个情况是一旦有较大值出现, 则不用全部遍历, 需要对比左右子树返回的值大小.

/**
 * 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(NULL==root)return -1;
        return traverse(root,root->val);
    }
    
    int traverse(TreeNode *root, int m)
    {
        if(NULL==root)return -1;
        if(root->val>m)return root->val;
        
        int l=traverse(root->left, m);
        int r=traverse(root->right, m);
        if(l!=-1&&r!=-1)return min(l,r);
        return max(l,r);
    }
};