Construct Binary Tree From Preorder and Inorder Traversal

 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     TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
13         if(preorder.size() == 0) return NULL;
14         return build(preorder, 0, preorder.size() - 1, inorder, 0, inorder.size() - 1);
15     }
16     TreeNode* build(vector<int>& preorder, int preLeft, int preRight, vector<int>& inorder, int inLeft, int inRight){
17         if(preLeft > preRight) return NULL; //when will this case appear
18         
19         TreeNode* root = new TreeNode(preorder[preLeft]);
20         if(preLeft == preRight) return root;
21         
22         //find the position of root(preorder[preLeft]) in inorder sequence
23         int index = 0;
24         for(; index < inorder.size(); index++){
25             if(inorder[inLeft + index] == preorder[preLeft]) break;
26         }
27         
28         root->left = build(preorder, preLeft + 1, preLeft + index, inorder, inLeft, inLeft + index - 1);
29         root->right = build(preorder, preLeft + index + 1, preRight, inorder, inLeft + index + 1, inRight);
30         
31         return root;
32     }
33 };