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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.

• Java
• C++
• Python
• Go
• TypeScript

• /**
   * 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;
  }
  
  

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