Two elements of a binary search tree (BST) are swapped by mistake.
Recover the tree without changing its structure.
Note:
confused what "{1,#,2,3}" means? >
read more on how binary tree is serialized on OJ.
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution
{
public:
void recoverTree(TreeNode *root)
{
if(root == NULL)
return;
vector<TreeNode*> res;
pre_search(root,res);
if(res.size() == 1)
return;
bool flag = false;
vector<int> temp;
for(int i=1; i<res.size(); i++)
{
if(res[i]->val <= res[i-1]->val)
{
if(flag == false)
{
temp.push_back(i-1);
flag = true;
}
else
{
temp.push_back(i);
flag = false;
}
}
}
int n = temp.size();
if(n == 1)
{
int pre = temp[0];
int t = res[pre]->val;
res[pre]->val = res[pre+1]->val;
res[pre+1]->val = t;
}
else
{
int pre = temp[0];
int end = temp[1];
int t = res[pre]->val;
res[pre]->val = res[end]->val;
res[end]->val = t;
}
return ;
}
void pre_search(TreeNode* root, vector<TreeNode*>& res)
{
if(root == NULL)
return;
pre_search(root->left,res);
res.push_back(root);
pre_search(root->right,res);
}
};LeetCode--Recover Binary Search Tree
原文:http://blog.csdn.net/shaya118/article/details/42695763