[LeetCode Road] Minimum Depth of Binary Tree - Solution/C++

111. Minimum Depth of Binary Tree

Question:

Given a binary tree, find its minimum depth.

The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node.

Note: A leaf is a node with no children.

Example:

Input: root = [3,9,20,null,null,15,7]
Output: 2

Input: root = [2,null,3,null,4,null,5,null,6]
Output: 5

Constraints:

The number of nodes in the tree is in the range [0, 10^5].

-1000 <= Node.val <= 1000

Source code:

Version 1

Idea:
We can sum up the rules of the binary tree.
First: If the left subnode and right subnode are exist of leaf node, it can be calculated each depths and use min() to get minimum depth.
Seconde: If there is NULL either left subnode or right subnode of leaf node, we must use max() to get the maxixmum depth, because "A leaf is a node with no children" based on the question. E.g., if left subnode is NULL, a leaf can't be calculated as 1 depth.

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int minDepth(TreeNode* root) {
        if(!root) return 0;
        if(root->left && root->right){
            return min(minDepth(root->left), minDepth(root->right)) + 1;
        }else{
            return max(minDepth(root->left), minDepth(root->right)) + 1;
        }
    }
};