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     struct TreeNode* reConstructBinaryTree(vector<int> pre,vector<int> in) {
13         if (pre.size() != 0 && in.size() != 0 && pre.size() == in.size())
14             {
15                 return build(pre,in,0,pre.size()-1,0,in.size()-1);
16             }
17         else
18             {
19                 return NULL;
20             }
21         
22     }
23     
24     struct TreeNode* build(vector<int> pre,vector<int> in,int l1,int r1,int l2,int r2)
25     {
26     if(l1<=r1)
27     {
28         int i=l2;
29         while(pre[l1]!=in[i])
30             i++;
31         TreeNode* T= new TreeNode(in[i]);
32         T->left=build(pre,in,l1+1,l1+i-l2,l2,i-1);
33         T->right=build(pre,in,l1+i-l2+1,r1,i+1,r2);
34         return T;
35     }
36     else return NULL;
37     }
38 };