LeetCode: Minimum Depth of Binary Tree

简单题，少数次过

 1 /**

 2  * Definition for binary tree

 3  * struct TreeNode {

 4  *     int val;

 5  *     TreeNode *left;

 6  *     TreeNode *right;

 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}

 8  * };

 9  */

10 class Solution {

11 public:

12     int dfs(TreeNode *root, int dep) {

13         if (!root) return dep;

14         if (root->left && root->right) return min(dfs(root->left, dep+1), dfs(root->right, dep+1));

15         if (!root->left && root->right) return dfs(root->right, dep+1);

16         if (root->left && !root->right) return dfs(root->left, dep+1);

17         if (!root->left && !root->right) return dep+1;

18     }

19     int minDepth(TreeNode *root) {

20         // Start typing your C/C++ solution below

21         // DO NOT write int main() function

22         return dfs(root, 0);

23     }

24 };

 C#, 没有return 0会有编译错误，C#的要求比C++的严格，但是功能又不全。。太废。。

 1 /**

 2  * Definition for a binary tree node.

 3  * public class TreeNode {

 4  *     public int val;

 5  *     public TreeNode left;

 6  *     public TreeNode right;

 7  *     public TreeNode(int x) { val = x; }

 8  * }

 9  */

10 public class Solution {

11     public int MinDepth(TreeNode root) {

12         return dfs(root, 0);

13     }

14     public int dfs(TreeNode root, int dep) {

15         if (root == null) return dep;

16         if (root.left != null && root.right != null) return Math.Min(dfs(root.left, dep+1), dfs(root.right, dep+1));

17         if (root.left == null && root.right != null) return dfs(root.right, dep+1);

18         if (root.left != null && root.right == null) return dfs(root.left, dep+1);

19         if (root.left == null && root.right == null) return dep + 1;

20         return 0;

21     }

22 }

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