题干：

代码：

/**
 * 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) {
        queue<TreeNode*>que;
        if(root != NULL) que.push(root);
        int depth = 0;
        while(!que.empty()){
            depth++;
            int size = que.size();
            for(int i = 0; i < size; i++){
                TreeNode* node = que.front();
                que.pop();
                if(node -> left) que.push(node -> left);
                if(node -> right) que.push(node -> right);
                if(!node -> left && !node -> right) return depth;//找不到左右节点就直接返回深度
            }
        }
        return depth;

    }
};