4、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> in) {
13         int n = pre.size();//确定节点数目
14         if(n <= 0)
15             return NULL;
16         else
17             return reConstructBinaryTree2(pre,0,n-1,in,0,n-1);
18     }
19     TreeNode* reConstructBinaryTree2(vector<int> pre,int preStart,int preEnd,vector<int> in,int inStart,int inEnd)
20     {
21         if(preStart > preEnd || inStart > inEnd)
22             return NULL;
23         TreeNode *root = new TreeNode(pre[preStart]);
24         for(int i = inStart; i <= inEnd; i++)
25         {
26             if(in[i] == pre[preStart])//在中序遍历中找到根节点
27             {
28                 root->left = reConstructBinaryTree2(pre,preStart+1,preStart+i-inStart,in,inStart,i-1);
29                 root->right = reConstructBinaryTree2(pre,preStart+i-inStart+1,preEnd,in,i+1,inEnd);
30             }
31         }
32         return root;
33          
34     }
35 };