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：

1     TreeNode *buildTree(vector
     
       &preorder, vector
      
        &inorder) { 2         // IMPORTANT: Please reset any member data you declared, as 3         // the same Solution instance will be reused for each test case. 4         int preIdx=0; 5         return buildTree(preorder,inorder,&preIdx,0,inorder.size()); 6     } 7     TreeNode* buildTree(vector
       
        & preorder,vector
        
         & inorder,int* preIdx,int inIdx,int inLen){ 8         if(preorder.size()==*preIdx||inLen==0) return NULL; 9         TreeNode* root = (TreeNode*)malloc(sizeof(TreeNode));10         root->val = preorder[*preIdx];11         root->left=NULL;12         root->right=NULL;13         int i=inIdx;14         while(i-inIdx
         
          val) break;16             i++;17         }18         if(i-inIdx>0&&*preIdx
          
           left=buildTree(preorder,inorder,&(++*preIdx),inIdx,i-inIdx);20 if(inLen-(i-inIdx)-1>0&&*preIdx
           
            right=buildTree(preorder,inorder,&(++*preIdx),i+1,inLen-(i-inIdx)-1);22 return root;23 }