construct-binary-tree-from-preorder-and-inorder-traversal——前序和中序求二叉树

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

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

 /**
  * Definition for binary tree
  * struct TreeNode {
  *     int val;
  *     TreeNode *left;
  *     TreeNode *right;
  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
  * };
  */
 class Solution {
 public:
     TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {
         if(preorder.size()!=inorder.size()) return NULL;
         return build(preorder,inorder,,preorder.size()-,,inorder.size()-);
     }
 
     TreeNode *build(vector<int> &preorder, vector<int> &inorder,int s1,int e1,int s2,int e2){
         if(s1>e1||s2>e2) return NULL;
         if(s1==e1) return new TreeNode(preorder[s1]);
         TreeNode *res=new TreeNode(preorder[s1]);
         int tmp=preorder[s1];
         int index=;
         while((s2+index)<=e2){
             if(inorder[s2+index]==tmp)
                 break;
             index++;
         }
         res->left=build(preorder,inorder,s1+,s1+index,s2,s2+index-);
         res->right=build(preorder,inorder,s1+index+,e1,s2+index+,e2);
         return res;
     }
 };