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
 
 /**
* 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) {
vector<TreeNode *> nodes;
dfs(root,nodes);
int index1=-;
int index2=-;
for(int i=;i<nodes.size();i++)
{
if(nodes[i]->val<nodes[i-]->val)
{
if(index1==-) index1=i-;
index2=i;
}
} int tmp=nodes[index1]->val;
nodes[index1]->val=nodes[index2]->val;
nodes[index2]->val=tmp;
return; } void dfs(TreeNode *root,vector<TreeNode *> &nodes)
{
if(root==NULL) return;
dfs(root->left,nodes);
nodes.push_back(root);
dfs(root->right,nodes);
}
};
 
 
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

 /**
* 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) { TreeNode *firstError=NULL;
TreeNode *secondError=NULL;
TreeNode *pre=NULL; dfs(root,pre,firstError,secondError); int tmp=firstError->val;
firstError->val=secondError->val;
secondError->val=tmp;
} void dfs(TreeNode *root,TreeNode *&pre,TreeNode *&firstError,TreeNode *&secondError)
{
if(root==NULL) return; dfs(root->left,pre,firstError,secondError); if(pre!=NULL&&pre->val>root->val)
{
if(firstError==NULL) firstError=pre;
secondError=root;
}
pre=root; dfs(root->right,pre,firstError,secondError);
}
};

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