111. Minimum Depth of Binary Tre

解法一：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 == NULL) return 0;
        int left = minDepth(root->left);
        int right = minDepth(root->right);    
        if(root->left == NULL) return 1 + right;
        if(root->right == NULL) return 1 + left;
        return 1 + min(left, right);
    }
};

解法二：bfs
注意的是：如何一层记一个数，用for循环来遍历q;

/**
 * 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 == NULL) return 0;
        queue<TreeNode*> q;
        q.push(root);
        int depth = 0;
        while(!q.empty()){
            int k = q.size();
            depth++;
            for(int i = 0; i < k; i++){
                TreeNode* cur = q.front();
                q.pop();
                if(cur->left == NULL && cur->right == NULL) return depth;
                if(cur->left) q.push(cur->left);
                if(cur->right) q.push(cur->right);
            }
        }
        return 0;
    }
};