一:题目

二：上码

/*** 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) {//思路:当我们遍历其（左右）孩子的节点为空的时候,这时候就到了叶节点，也就是最小的时候queue<TreeNode*>q;if(root) q.push(root);int minDepth = 0;while(!q.empty()) {int size = q.size();minDepth++;for(int i = 0; i < size; i++) {TreeNode* node = q.front();q.pop();if(node->left) q.push(node->left);if(node->right) q.push(node->right);if(node->left == NULL && node->right == NULL) return minDepth;}}return minDepth;}
};