Leetcode每日一题：111.minimum-depth-of-binary-tree(二叉树的最小深度)

思路：这道题的话比较容易想通的，就利用队列进行层次遍历，每下一层就将高度+1，然后在每一层的遍历中，如果发现某个结点没有孩子结点，那么它就是高度最小的叶子结点, 由此得到这棵树的最小深度；

struct TreeNode
{int val;TreeNode *left;TreeNode *right;TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};int minDepth(TreeNode *root)
{if (root == NULL)  //该树为空 直接返回0return 0;queue<TreeNode *> q;q.push(root);   //把根结点放入队列int result = 0;while (!q.empty()){queue<TreeNode *> qq;  //辅助层次队列result++;while (!q.empty()){TreeNode *now = q.front();q.pop();if (now->left == NULL && now->right == NULL) //如果当前结点没有孩子结点，直接就作为深度最小的叶子节点，返回深度值return result;if (now->left)//如果不是的话，就把它非空的孩子结点放入队列中{qq.push(now->left);}if (now->right){qq.push(now->right);}}q = qq;}return result;
}

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