重建二叉树

/**
* Definition for binary tree
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
struct TreeNode* reConstructBinaryTree(vector<int> pre,vector<int> in) {

vector<int> preLeft,preRight,inLeft,inRight;
int length=pre.size();
if(length<=0)
return NULL;
struct TreeNode* tree=new TreeNode(pre[0]);//把节点值放进去,此处必须写作构造函数形式
//如果不为tree指针指定地址，会成为一个野指针，造成泄漏
int indexRootPre;//根节点在中序遍历的位置
for(int i=0;i<in.size();i++){
if(in[i]==pre[0]){
indexRootPre=i;
break;
}
}
for(int i=0;i<length;i++){//分割前序遍历分为左右两个子树
if(i<indexRootPre)
inLeft.push_back(in[i]);
if(i>indexRootPre)
inRight.push_back(in[i]);
}
for(int i=1;i<length;i++){//0已经作为根节点，所以从1开始分割中序遍历分为左右两个子树
if(i<=indexRootPre)
preLeft.push_back(pre[i]);
if(i>indexRootPre)
preRight.push_back(pre[i]);
}
tree->left=reConstructBinaryTree(preLeft,inLeft);//构造左子树
tree->right=reConstructBinaryTree(preRight,inRight);//构造右字树
return tree;
}
};