Recover Binary Search Tree
Problem Statement :
You are given the root of a binary search tree (BST), where the values of exactly two nodes of the tree were swapped by mistake. Recover the tree without changing its structure. Example 1: Input: root = [1,3,null,null,2] Output: [3,1,null,null,2] Explanation: 3 cannot be a left child of 1 because 3 > 1. Swapping 1 and 3 makes the BST valid. Example 2: Output: [2,1,4,null,null,3] Explanation: 2 cannot be in the right subtree of 3 because 2 < 3. Swapping 2 and 3 makes the BST valid. Constraints: The number of nodes in the tree is in the range [2, 1000]. -231 <= Node.val <= 231 - 1
Solution :
Solution in C :
struct TreeNode * inorder(struct TreeNode * root, struct TreeNode ** p1, struct TreeNode ** p2, int d);
void recoverTree(struct TreeNode* root){
struct TreeNode * p1 = NULL;
struct TreeNode * p2 = NULL;
int l = 0, r = 1;
inorder(root, &p1, &p2, 0);
p1->val = p1->val + p2->val;
p2->val = p1->val - p2->val;
p1->val = p1->val - p2->val;
}
struct TreeNode * inorder(struct TreeNode * root, struct TreeNode ** p1, struct TreeNode ** p2, int d){
if (root == NULL)
return NULL;
struct TreeNode * cur = NULL;
cur = inorder(root->left, p1, p2, 0);
if (cur != NULL){
if (cur->val > root->val){
if (*p1 == NULL){
*p1 = cur;
*p2 = root;
}
else
*p2 = root;
}
}
int in_left = 0;
if (*p1 != NULL)
in_left = 1;
cur = inorder(root->right, p1, p2, 1);
if (cur != NULL){
if (root->val > cur->val){
if (*p1 == NULL){
*p1 = root;
*p2 = cur;
}
if (in_left == 1)
*p2 = cur;
else
*p1 = root;
}
}
if (d == 0){
while (root->right != NULL)
root = root->right;
}
else{
while (root->left != NULL)
root = root->left;
}
return root;
}
Solution in C++ :
class Solution {
public:
TreeNode* first = NULL, *second = NULL;
void inorder(TreeNode* root, TreeNode* &prev){
if(!root) return;
inorder(root->left, prev);
if(root->val<prev->val){
if(!first)
{first = prev;second=root;}
else second=root;
}
prev=root;
inorder(root->right, prev);
}
void recoverTree(TreeNode* root) {
TreeNode* prev=new TreeNode(INT_MIN);
inorder(root, prev);
if(first) swap(first->val, second->val);
}
};
Solution in Java :
class Solution {
public void recoverTree(TreeNode root) {
if(root == null) return;
TreeNode curr = root;
TreeNode num1 = null;
TreeNode num2 = null;
TreeNode prev =null;
while(curr!=null){
if(curr.left==null){
if(prev!=null && prev.val>curr.val){
if(num1==null && num2==null){
num1 = prev;
num2 = curr;
}
else{
num2 = curr;
}
}
prev = curr;
curr = curr.right;
}
else{
TreeNode temp = curr.left;
while(temp.right!=null && temp.right!=curr){
temp = temp.right;
}
if(temp.right==null){
temp.right = curr;
curr = curr.left;
}
else{
if(prev!=null && prev.val>curr.val){
if(num1==null && num2==null){
num1 = prev;
num2 = curr;
}
else{
num2 = curr;
}
}
prev = curr;
temp.right = null;
curr = curr.right;
}
}
}
int temp = num1.val;
num1.val = num2.val;
num2.val = temp;
}
}
Solution in Python :
class Solution:
def recoverTree(self, root: Optional[TreeNode]) -> None:
prev, first, second = None, None, None
def inorder(node):
nonlocal prev, first, second
if node.left:
inorder(node.left)
if prev and first is None and prev.val>node.val:
first = prev
if prev and first is not None and prev.val>node.val:
second = node
prev = node
if node.right:
inorder(node.right)
inorder(root)
tmp = first.val
first.val = second.val
second.val = tmp
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