Given the root of a binary search tree, rearrange the tree in in-order so that the leftmost node in the tree is now the root of the tree, and every node has no left child and only one right child.
Example 1:
Input: root = [5,3,6,2,4,null,8,1,null,null,null,7,9] Output: [1,null,2,null,3,null,4,null,5,null,6,null,7,null,8,null,9]
Example 2:
Input: root = [5,1,7] Output: [1,null,5,null,7]
Constraints:
- The number of nodes in the given tree will be in the range
[1, 100]. 0 <= Node.val <= 1000
Related Topics:
Tree, Depth-first Search, Recursion
// OJ: https://leetcode.com/problems/increasing-order-search-tree
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(H)
class Solution {
private:
TreeNode *prev;
void inorder(TreeNode *root) {
if (!root) return;
inorder(root->left);
root->left = NULL;
prev->right = root;
prev = root;
inorder(root->right);
}
public:
TreeNode* increasingBST(TreeNode* root) {
TreeNode head;
prev = &head;
inorder(root);
return head.right;
}
};// OJ: https://leetcode.com/problems/increasing-order-search-tree/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(H)
class Solution {
pair<TreeNode*, TreeNode*> dfs(TreeNode* root) {
TreeNode *head = root, *tail = root;
if (root->left) {
auto [leftHead, leftTail] = dfs(root->left);
head = leftHead;
leftTail->right = root;
root->left = NULL;
}
if (root->right) {
auto [rightHead, rightTail] = dfs(root->right);
root->right = rightHead;
tail = rightTail;
}
return { head, tail };
}
public:
TreeNode* increasingBST(TreeNode* root) {
return dfs(root).first;
}
};