# 897. Increasing Order Search Tree

## Description

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

## Solutions

Solution 1: DFS In-order Traversal

We define a virtual node $dummy$, initially the right child of $dummy$ points to the root node $root$, and a pointer $prev$ points to $dummy$.

We perform an in-order traversal on the binary search tree. During the traversal, each time we visit a node, we point the right child of $prev$ to it, then set the left child of the current node to null, and assign the current node to $prev$ for the next traversal.

After the traversal ends, the original binary search tree is modified into a singly linked list with only right child nodes. We then return the right child of the virtual node $dummy$.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the number of nodes in the binary search tree.

• /**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode() {}
*     TreeNode(int val) { this.val = val; }
*     TreeNode(int val, TreeNode left, TreeNode right) {
*         this.val = val;
*         this.left = left;
*         this.right = right;
*     }
* }
*/
class Solution {
private TreeNode prev;
public TreeNode increasingBST(TreeNode root) {
TreeNode dummy = new TreeNode(0, null, root);
prev = dummy;
dfs(root);
return dummy.right;
}

private void dfs(TreeNode root) {
if (root == null) {
return;
}
dfs(root.left);
prev.right = root;
root.left = null;
prev = root;
dfs(root.right);
}
}

• /**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode() : val(0), left(nullptr), right(nullptr) {}
*     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
*     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
TreeNode* increasingBST(TreeNode* root) {
TreeNode* dummy = new TreeNode(0, nullptr, root);
TreeNode* prev = dummy;
function<void(TreeNode*)> dfs = [&](TreeNode* root) {
if (!root) {
return;
}
dfs(root->left);
prev->right = root;
root->left = nullptr;
prev = root;
dfs(root->right);
};
dfs(root);
return dummy->right;
}
};

• # Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
def increasingBST(self, root: TreeNode) -> TreeNode:
def dfs(root):
if root is None:
return
nonlocal prev
dfs(root.left)
prev.right = root
root.left = None
prev = root
dfs(root.right)

dummy = prev = TreeNode(right=root)
dfs(root)
return dummy.right


• /**
* Definition for a binary tree node.
* type TreeNode struct {
*     Val int
*     Left *TreeNode
*     Right *TreeNode
* }
*/
func increasingBST(root *TreeNode) *TreeNode {
dummy := &TreeNode{Val: 0, Right: root}
prev := dummy
var dfs func(root *TreeNode)
dfs = func(root *TreeNode) {
if root == nil {
return
}
dfs(root.Left)
prev.Right = root
root.Left = nil
prev = root
dfs(root.Right)
}
dfs(root)
return dummy.Right
}

• /**
* Definition for a binary tree node.
* class TreeNode {
*     val: number
*     left: TreeNode | null
*     right: TreeNode | null
*     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
*         this.val = (val===undefined ? 0 : val)
*         this.left = (left===undefined ? null : left)
*         this.right = (right===undefined ? null : right)
*     }
* }
*/

function increasingBST(root: TreeNode | null): TreeNode | null {
const dummy = new TreeNode((right = root));
let prev = dummy;
const dfs = (root: TreeNode | null) => {
if (!root) {
return;
}
dfs(root.left);
prev.right = root;
root.left = null;
prev = root;
dfs(root.right);
};
dfs(root);
return dummy.right;
}