【树】105. 从前序与中序遍历序列构造二叉树

 1 /**
 2  * Definition for a binary tree node.
 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     
13     TreeNode *buildTreeHelper(vector<int>& preorder, int ps, int pe, vector<int>& inorder, int is, int ie)
14     {
15         if ((pe - ps < 0)  || (pe - ps) != (ie - is))
16         {
17             return NULL;
18         } 
19 
20         TreeNode *root = new TreeNode(preorder[ps]);
21 
22         int index = -1;
23         for (int i = is; i <= ie; i++)
24         {
25             if (inorder[i] == root->val)
26             {
27                 index = i;
28                 break;
29             }
30         }
31 
32         if (index == -1)
33         {
34             return root;
35         }
36 
37         int leftprelen = index - is;
38         root->left = buildTreeHelper(preorder, ps + 1, ps + leftprelen, inorder, is, index - 1);
39         root-> right = buildTreeHelper(preorder, ps + leftprelen + 1, pe, inorder, index + 1, ie);
40 
41         return root;
42     }
43 
44     TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) 
45     {
46         return buildTreeHelper(preorder, 0, preorder.size() - 1, inorder, 0, inorder.size() - 1);    
47     }
48 };