Construct Binary Tree from Preorder and Inorder Traversal

题目：

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

通过先序遍历和中序遍历构造二叉树。大概思路是递归的构造，因为先序遍历总是先访问根结点，所以很容易从先序列表中得到根，位于该结点右侧的就是子树，再由这个根结点从中序列表中找到，位于该值左侧的就是左子树，右侧的即为又子树。然后分别用同样的方法构建左、右子树，直到构造完成。代码：

    TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        int preIdx=0;
        return buildTree(preorder,inorder,&preIdx,0,inorder.size());
    }
    TreeNode* buildTree(vector<int>& preorder,vector<int>& inorder,int* preIdx,int inIdx,int inLen){
        if(preorder.size()==*preIdx||inLen==0) return NULL;
        TreeNode* root = (TreeNode*)malloc(sizeof(TreeNode));
        root->val = preorder[*preIdx];
        root->left=NULL;
        root->right=NULL;
        int i=inIdx;
        while(i-inIdx<inLen){
            if(inorder[i]==root->val) break;
            i++;
        }
        if(i-inIdx>0&&*preIdx<preorder.size())//no left child
            root->left=buildTree(preorder,inorder,&(++*preIdx),inIdx,i-inIdx);
        if(inLen-(i-inIdx)-1>0&&*preIdx<preorder.size())//no right child
            root->right=buildTree(preorder,inorder,&(++*preIdx),i+1,inLen-(i-inIdx)-1);
        return root;
    }