<span>重建二叉树</span>

 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     struct TreeNode* reConstructBinaryTree(vector<int> pre,vector<int> in) {
13         if(pre.empty() || in.empty())
14             return NULL;
15         return ConstructBinaryTree(pre.begin(),pre.end()-1,in.begin(),in.end()-1);
16     }
17     struct TreeNode* ConstructBinaryTree
18     (vector<int>::iterator startPre,vector<int>::iterator endPre,
19      vector<int>::iterator startIn,vector<int>::iterator endIn){
20         int rootValue = *startPre;
21         TreeNode* root = new TreeNode(rootValue);
22         if(startPre == endPre){
23             if(startIn == endIn && *startPre == *startIn)
24                 return root;
25             else
26                 return NULL;
27         }
28         vector<int>::iterator rootInOrder = startIn;
29         while(rootInOrder <= endIn && *rootInOrder != rootValue)
30             ++rootInOrder;
31         
32         if(rootInOrder == endIn && *rootInOrder != rootValue)
33             return NULL;
34         
35         int leftLength = rootInOrder - startIn;
36         vector<int>::iterator leftPreEnd = startPre + leftLength;
37         if(leftLength>0)
38             root->left = ConstructBinaryTree(startPre+1,leftPreEnd,startIn,rootInOrder-1);
39         if(leftLength < endIn - startIn)
40             root->right = ConstructBinaryTree(leftPreEnd+1,endPre,rootInOrder+1,endIn);
41         
42         return root;
43     }
44 };