Leetcode144: Construct 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.

/** * 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:    TreeNode* dfs(vector<int>& preorder, int prestart, int preend, vector<int>& inorder, int instart, int inend)    {        if(prestart > preend)            return NULL;                TreeNode* root = new TreeNode(0);        root->val = preorder.at(prestart);        if(prestart == preend)            return root;        int i;        for(i = 0; i <= (inend-instart); i++)        {            if(inorder.at(instart+i) == root->val)                break;        }        root->left = dfs(preorder, prestart+1, prestart+i, inorder, instart, instart+i-1);        root->right = dfs(preorder, prestart+i+1, preend, inorder, instart+i+1, inend);                return root;            }        TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {        if(preorder.size()<=0 || preorder.size()!=inorder.size())            return NULL;        TreeNode* root = new TreeNode(0);        root = dfs(preorder, 0, preorder.size()-1, inorder, 0, inorder.size()-1);        return root;    }};