LeetCode: Construct Binary Tree from Inorder and Postorder Traversal

第一次只用4个系数写dfs怎么写都不对，后来看了网上的答案发现写右子树的时候开始点不一样的。。。所以要用5个系数

 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> &inorder, vector<int> &postorder, int inbeg, int postbeg, int len) {

13         if (!len) return NULL;

14         TreeNode *tmp = new TreeNode(postorder[postbeg+len-1]);

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             }

21         }

22         tmp->left = dfs(inorder, postorder, inbeg, postbeg, leftlen);

23         tmp->right = dfs(inorder, postorder, inbeg+leftlen+1, postbeg+leftlen, rightlen);

24         return tmp;

25     }

26     TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) {

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

28         // DO NOT write int main() function

29         if (inorder.size() != postorder.size() || inorder.size() == 0) return NULL;

30         int len = inorder.size();

31         return dfs(inorder, postorder, 0, 0, len);

32     }

33 };

 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[] inorder, int[] postorder) {

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

13         int len = inorder.Length;

14         return dfs(inorder, postorder, 0, 0, ref len);

15     }

16     TreeNode dfs(int[] inorder, int[] postorder, int inbeg, int postbeg, ref int len)

17     {

18         if (len == 0) return null;

19         TreeNode tmp = new TreeNode(postorder[postbeg+len-1]);

20         int leftlen = 0;

21         int rightlen = 0;

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

23         {

24             if (tmp.val == inorder[i])

25             {

26                 leftlen = i - inbeg;

27                 rightlen = len - leftlen - 1;

28             }

29         }

30         tmp.left = dfs(inorder, postorder, inbeg, postbeg, ref leftlen);

31         tmp.right = dfs(inorder, postorder, inbeg+leftlen+1, postbeg+leftlen, ref rightlen);

32         return tmp;

33     }

34 }

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