【leetcode】Construct Binary Tree from Preorder and Inorder Traversal

Question：  

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

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

Anwser 1 ：      

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    TreeNode *build(vector<int> &preorder, vector<int> &inorder, int sPre, int ePre, int sMid, int eMid) {        if (sPre > ePre || sMid > eMid) return NULL;                int pivot = preorder[sPre];                int pos;        for (pos = sMid; pos <= eMid; pos++) {            if (inorder[pos] == pivot)                break;        }                TreeNode *parent = new TreeNode(pivot);                parent->left  = build(preorder, inorder, sPre + 1, sPre + (pos - sMid), sMid, pos - 1);        parent->right = build(preorder, inorder, sPre + (pos - sMid) + 1, ePre, pos + 1, eMid);                return parent;    }        TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        return build(preorder, inorder, 0, preorder.size() - 1, 0, inorder.size() - 1);    }};

Anwser 2 ：   

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        int size = preorder.size();        if(size == 0) return NULL;                return BuildBT(preorder, inorder, 0, 0, size-1);    }    TreeNode *BuildBT(vector<int> &preorder, vector<int> &inorder, int pos, int start, int end){        if(start > end ) return NULL;                  int j = std::find(inorder.begin() + start, inorder.end() + end, preorder[pos]) - inorder.begin();        TreeNode *root = new TreeNode(inorder[j]);        root->left = BuildBT(preorder, inorder, pos+1, start, j-1);        root->right = BuildBT(preorder, inorder, j+pos-start+1, j+1, end); // j+pos-start+1 is the relative position in the preorder array of the  next right child                   return root;    }};