LeetCode|Second Minimum Node In a Binary Tree

最新推荐文章于 2025-07-08 14:36:18 发布

DouMiaoO_Oo

最新推荐文章于 2025-07-08 14:36:18 发布

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分类专栏： LeetCode 文章标签： LeetCode 二叉树

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本文介绍了一种算法，用于找到特殊二叉树中第二小的节点值。该算法通过递归遍历树并比较节点值来寻找目标值，优化后的实现效率高，适用于寻找特定值的问题。

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Second Minimum Node In a Binary Tree

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.

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.

/**
 * 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) {
        fir = root->val;
        sec = -1;
        return helper(root);
    }
private:
    int fir, sec;
    int helper(TreeNode* root){
        if(root == NULL) return -1;
        if(root->val > fir) return root->val;
        if(root->left == NULL && root->right == NULL){
            return -1;
        }
        if(root->left != NULL && root->right == NULL){
            return helper(root->left);
        }
        if(root->left == NULL && root->right != NULL){
            return helper(root->right);
        }
        
        if(root->left->val == fir || root->right->val == fir){
            // 只要有一棵子树等于fir就得全部搜索
            int left = helper(root->left);
            int right = helper(root->right);
            if(left == fir && right == fir) return -1;
            if(left > fir && right > fir) return min(left, right);
            return max(left, right);
        } return min(root->left->val, root->right->val);
        
        
    }
};

不枉我优化了一下… 提交的结果是：
Runtime: 4 ms, faster than 100.00% of C++ online submissions for Second Minimum Node In a Binary Tree.
Memory Usage: 8.7 MB, less than 58.65% of C++ online submissions for Second Minimum Node In a Binary Tree.