【leetcode】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.

 

class Solution {
public:
    TreeNode* createTree(vector<int>& preorder,int prestart,int preend,vector<int>& inorder,int instart,int inend)
    {
        if(instart>inend) return NULL;

        int root=preorder[prestart];
        int index;

        for(int i=instart;i<=inend;i++)
        {
            if(inorder[i]==root)
            {
                index=i;
                break;
            }
        }

        int len=index-instart;
        TreeNode * left=createTree(preorder,prestart+1,prestart+len,inorder,instart,index-1);
        //TreeNode * left=createTree(preorder,prestart+1,prestart+len-1,inorder,instart,index-1);
        //TreeNode * right=createTree(preorder,prestart+len,preend-1,inorder,index+1,inend);
        TreeNode * right=createTree(preorder,prestart+len+1,preend,inorder,index+1,inend);
        TreeNode *node=new TreeNode(root);
        node->left=left;
        node->right=right;

        return node;
    }


    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        if(preorder.size()==0) return NULL;

        TreeNode* head=createTree(preorder,0,preorder.size()-1,inorder,0,inorder.size()-1);
        return head;
    }
};