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?
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
private:
void swap(int& a,int&b)
{
int tmp=a;
a=b;
b=tmp;
}
TreeNode* findlarge(TreeNode* root,int val)
{
if(root==NULL) return NULL;
TreeNode* result=NULL;
if(root->val>val)
{
val=root->val;
result=root;
}
TreeNode* p1=findlarge(root->left,val);
TreeNode* p2=findlarge(root->right,val);
if(p1!=NULL) result=p1;
if(p2!=NULL)
{
if(result==NULL) result=p2;
else
if(result->val<p2->val) result=p2;
}
return result;
}
TreeNode* findsmall(TreeNode* root,int val)
{
if(root==NULL) return NULL;
TreeNode* result=NULL;
if(root->val<val)
{
val=root->val;
result=root;
}
TreeNode* p1=findsmall(root->left,val);
TreeNode* p2=findsmall(root->right,val);
if(p1!=NULL) result=p1;
if(p2!=NULL)
{
if(result==NULL) result=p2;
else
if(result->val>p2->val) result=p2;
}
return result;
}
public:
void recoverTree(TreeNode *root)
{
if(root==NULL) return;
TreeNode* left=findlarge(root->left,root->val);
TreeNode* right=findsmall(root->right,root->val);
if(left!=NULL && right!=NULL)
{
swap(left->val,right->val);
return;
}
if(right!=NULL)
{
swap(root->val,right->val);
return;
}
if(left!=NULL)
{
swap(root->val,left->val);
return;
}
recoverTree(root->left);
recoverTree(root->right);
}
};
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时间： 2024-12-14 01:41:39