解法1 递归

/**
 * 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 minDepth(TreeNode* root) {
        if(!root)   return 0;
        if(!root->left&&!root->right)   return 1;
        if(root->left&&root->right)     return 1+min(minDepth(root->left),minDepth(root->right));
        if(root->left)  return minDepth(root->left)+1;
        return minDepth(root->right)+1;
    }
};

解法2 BFS

class Solution {
public:

    int minDepth(TreeNode* root) {
        if(!root)   return 0;
        return bfs(root); 
    }

    int bfs(TreeNode * u){
        queue<pair<TreeNode *,int>> q;
        q.push(make_pair(u,1));   
        while(q.size()){
            auto t=q.front();
            q.pop();
            TreeNode * node=t.first;
            if(!node->left&&!node->right)
                return t.second;
            if(node->right) q.push(make_pair(node->right,t.second+1));
            if(node->left) q.push(make_pair(node->left,t.second+1));      

        }
        return -1;
    }

};