题解 | #牛群的最短路径#

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 *	TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 * };
 */
class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param root TreeNode类 
     * @return int整型
     */
    int minDepth(TreeNode* root) 
    {
        // write code here
        if (root == nullptr) 
        {
            return 0;
        }
        
        // Base case: leaf node
        if (root->left == nullptr && root->right == nullptr) 
        {
            return 1;
        }
        
        // Calculate depth of left and right subtrees
        int leftDepth = INT_MAX;
        int rightDepth = INT_MAX;
        
        if (root->left != nullptr) 
        {
            leftDepth = minDepth(root->left);
        }
        
        if (root->right != nullptr) 
        {
            rightDepth = minDepth(root->right);
        }
        
        // Return the minimum depth of the subtree + 1
        return min(leftDepth, rightDepth) + 1;
    }
};