[剑指Offer]重建二叉树

 1 /**
 2  * Definition for binary tree
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     TreeNode* reConstructBinaryTree(vector<int> pre,vector<int> vin) {
13       //left-close-right-open 
14       return RebuildBST(pre,vin,0,pre.size(),0,vin.size());
15       
16     }
17     TreeNode* RebuildBST(vector<int> pre,vector<int> vin,int preBegin,int preEnd,int inBegin,int inEnd)
18     {
19       if(preBegin >= preEnd || inBegin>=inEnd) return nullptr;
20       int rootCur = pre[preBegin];
21       TreeNode* root = new TreeNode(rootCur);
22       int rootPos = 0;
23       for(rootPos = inBegin;rootPos<inEnd;rootPos++)
24       {
25         if(vin[rootPos] == rootCur)
26         {
27           break;
28         }
29       }
30       root->left = RebuildBST(pre,vin,preBegin + 1, preBegin + rootPos - inBegin + 1 ,inBegin,rootPos);
31       root->right = RebuildBST(pre,vin,preBegin + rootPos - inBegin + 1, preEnd,rootPos + 1,inEnd);
32 
33       return root;
34     }
35 };