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

Note:

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

class Solution {public:    /* from Preorder and Inorder Traversal */    TreeNode* buildTree(vector
     
      & preorder, vector
      
       & inorder) {        return helper(preorder,0,preorder.size(),inorder,0,inorder.size());    }    TreeNode* helper(vector
       
        & preorder,int i,int j,vector
        
         & inorder,int ii,int jj)    {        // tree        8 4 5 3 7 3        // preorder    8 [4 3 3 7] [5]        // inorder     [3 3 4 7] 8 [5]        // 每次从 preorder 头部取一个值 mid，作为树的根节点        // 检查 mid 在 inorder 中 的位置，则 mid 前面部分将作为 树的左子树，右部分作为树的右子树        if(i >= j || ii >= j)            return NULL;        int mid = preorder[i];        auto f = find(inorder.begin() + ii,inorder.begin() + jj,mid);        int dis = f - inorder.begin() - ii;        TreeNode* root = new TreeNode(mid);        root -> left = helper(preorder,i + 1,i + 1 + dis,inorder,ii,ii + dis);        root -> right = helper(preorder,i + 1 + dis,j,inorder,ii + dis + 1,jj);        return root;    }};