LeetCode 105. Construct Binary Tree from Preorder and Inorder Traversal 由前序和中序遍历建立二叉树 C++

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

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

For example, given

preorder = [3,9,20,15,7]
inorder = [9,3,15,20,7]

Return the following binary tree:

    3
   / 
  9  20
    /  
   15   7

前序、中序遍历得到二叉树，可以知道每一次前序新数组的第一个数为其根节点。在中序遍历中找到根节点对应下标，根结点左边为其左子树，根节点右边为其右子树，再根据中序数组中的左右子树个数，找到对应的前序数组中的左右子树新数组。设每次在中序数组中对应的位置为i，则对应的新数组为：

前序新数组：左子树[pleft+1,pleft+i-ileft],右子树[pleft+i-ileft+1,pright]

中序新数组：左子树[ileft,i-1],右子树[i+1,iright]  C++

TreeNode* buildTree(vector<int>& preorder,int pleft,int pright,vector<int>& inorder,int ileft,int iright){
        if(pleft>pright||ileft>iright)
            return NULL;
        int i=0;
        for(i=ileft;i<=iright;i++){
            if(preorder[pleft]==inorder[i])
                break;
        }
        TreeNode* cur=new TreeNode(preorder[pleft]);
        cur->left=buildTree(preorder,pleft+1,pleft+i-ileft,inorder,ileft,i-1);
        cur->right=buildTree(preorder,pleft+i-ileft+1,pright,inorder,i+1,iright);
        return cur;
    }
    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        return buildTree(preorder,0,preorder.size()-1,inorder,0,inorder.size()-1);
    }