Leetcode671. Second Minimum Node In a Binary Tree

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本文介绍了一种解决特殊二叉树中寻找第二最小值的方法。通过递归遍历树结构，并利用两个变量记录最小值及次小值，最终输出所求值。如果不存在第二小值，则返回-1。

题目：

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

题目分析：

这是一棵树，首先考虑递归。然后要找到第二小的结点。

所以用一个min1, min2的结点去标记。

代码：

class Solution {
public:
    int min1 = INT32_MAX, min2 = INT32_MAX;
    int findSecondMinimumValue(TreeNode* root) {
        inorder(root);
        return min2 == INT32_MAX?-1:min2;
    }
    void inorder(TreeNode* root) {
        if (!root) return;
        if (root->val < min1) {
            min2 = min1;
            min1 = root->val;
        } else if (root->val > min1 && root->val < min2) {
            min2 = root->val;
        }
        inorder(root->left);
        inorder(root->right);
    }
};