leetcode 105 Constract Binary Tree from Preorder and Inorder Traversal

Given preorder and inorder traversal of a tree, construct the binary tree.

Note:

You may assume that duplicates do not exist in the tree.

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/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
void tree(TreeNode *&root, vector<int> &preorder, int &num, vector<int> &inorder, int left, int right) {
if(left > right) {
root = NULL;
return ;
}
root = new TreeNode(preorder[num++]);
int temp = preorder[num-1], i;

for(i=left; i <= right; i++) {
if(inorder[i]==temp) break;
}
tree(root->left, preorder, num, inorder, left, i-1);
tree(root->right, preorder, num, inorder, i+1, right);
}
TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {
TreeNode *root=NULL;
if(preorder.size()==0 || inorder.size()==0) return root;

int num = 0, i=0, left=0, right=inorder.size();
tree(root, preorder, num, inorder, left, right-1);
return root;
}
};