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 };

 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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posted @ 2013-03-19 15:21 ying_vincent 阅读(172) 评论(0) 收藏举报

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