LeetCode: Construct Binary Tree from Preorder and Inorder Traversa

跟inorder and postorder基本一样，一次过吧

 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     TreeNode *dfs(vector<int> &preorder, vector<int> &inorder, int prebeg, int inbeg, int len) {

13         if (!len) return NULL;

14         TreeNode *tmp = new TreeNode(preorder[prebeg]);

15         int leftlen, rightlen;

16         for (int i = inbeg; i < inbeg+len; i++) {

17             if (tmp->val == inorder[i]) {

18                 leftlen = i - inbeg;

19                 rightlen = len - leftlen - 1;

20                 break;

21             }

22         }

23         tmp->left = dfs(preorder, inorder, prebeg+1, inbeg, leftlen);

24         tmp->right = dfs(preorder, inorder, prebeg+leftlen+1, inbeg+leftlen+1, rightlen);

25         return tmp;

26     }

27     TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {

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

29         // DO NOT write int main() function

30         if (preorder.size() != inorder.size() || preorder.size() == 0) return NULL;

31         int len = preorder.size();

32         return dfs(preorder, inorder, 0, 0, len);

33     }

34 };

 C#　

 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 TreeNode BuildTree(int[] preorder, int[] inorder) {

12         if (preorder.Length == 0 || preorder.Length != inorder.Length) return null;

13         int len = preorder.Length;

14         return dfs(preorder, inorder, 0, 0, len);

15     }

16     TreeNode dfs(int[] preorder, int[] inorder, int prebeg, int inbeg, int len)

17     {

18         if (len == 0) return null;

19         TreeNode tmp = new TreeNode(preorder[prebeg]);

20         int leftlen = 0;

21         int rightlen = 0;

22         for (int i = inbeg; i < inbeg+len; i++) {

23             if (tmp.val == inorder[i]) {

24                 leftlen = i - inbeg;

25                 rightlen = len - leftlen - 1;

26                 break;

27             }

28         }

29         tmp.left = dfs(preorder, inorder, prebeg+1, inbeg, leftlen);

30         tmp.right = dfs(preorder, inorder, prebeg+leftlen+1, inbeg+leftlen+1, rightlen);

31         return tmp;

32     }

33 }

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