要求:此题正好和Maximum Depth of Binary Tree一题是相反的,即寻找二叉树的最小的深度值:从根节点到最近的叶子节点的距离。

结题思路:和找最大距离不同之处在于:找最小距离要注意(l<r)? l+1:r+1的区别应用,因为可能存在左右子树为空的情况,此时值就为0,但显然值是不为0的(只有当二叉树为空才为0),所以,在这里注意一下即可!

代码如下:

 struct TreeNode {
int val;
TreeNode *left;
TreeNode *right;
TreeNode(int x): val(x),left(NULL), right(NULL) {}
}; int minDepth(TreeNode *root)
{
if (NULL == root)
return ;
int l = minDepth(root->left);
int r = minDepth(root->right);
if (!l)
return r+;
if (!r)
return l+;
return (l<r)?l+1:r+1;
}

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