【leetcode】Recover Binary Search Tree

Recover Binary Search Tree

Two elements of a binary search tree (BST) are swapped by mistake.

Recover the tree without changing its structure.

Note:
A solution using O( n) space is pretty straight forward. Could you devise a constant space solution?

 

中序遍历解法O(n)space

 

 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     void recoverTree(TreeNode *root) {

13         vector<TreeNode *> nodes;

14         dfs(root,nodes);

15         int index1=-1;

16         int index2=-1;

17         for(int i=1;i<nodes.size();i++)

18         {

19             if(nodes[i]->val<nodes[i-1]->val)

20             {

21                 if(index1==-1) index1=i-1;

22                 index2=i;

23             }

24         }

25        

26         int tmp=nodes[index1]->val;

27         nodes[index1]->val=nodes[index2]->val;

28         nodes[index2]->val=tmp;

29         return;

30        

31     }     

32    

33     void dfs(TreeNode *root,vector<TreeNode *> &nodes)

34     {

35         if(root==NULL) return;

36         dfs(root->left,nodes);

37         nodes.push_back(root);

38         dfs(root->right,nodes);

39     }

40 };

 

 

 

直接在中序遍历中进行判断，需要记录遍历中的前一个元素

如果前一个元素比当前元素大，则说明存在错误

 

 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     void recoverTree(TreeNode *root) {

13     

14         TreeNode *firstError=NULL;

15         TreeNode *secondError=NULL;

16         TreeNode *pre=NULL;

17 

18         

19         dfs(root,pre,firstError,secondError);

20         

21         int tmp=firstError->val;

22         firstError->val=secondError->val;

23         secondError->val=tmp;

24     }

25     

26     void dfs(TreeNode *root,TreeNode *&pre,TreeNode *&firstError,TreeNode *&secondError)

27     {

28         if(root==NULL) return;

29         

30         dfs(root->left,pre,firstError,secondError);

31         

32         if(pre!=NULL&&pre->val>root->val)

33         {

34             if(firstError==NULL) firstError=pre;

35             secondError=root;

36         }

37         pre=root;

38         

39         dfs(root->right,pre,firstError,secondError);

40     }

41 };