Formatted question description: https://leetcode.ca/all/897.html

897. Increasing Order Search Tree (Easy)

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

Solution 1. In-order Traversal

// OJ: https://leetcode.com/problems/increasing-order-search-tree
// 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;
    }
};

Solution 2. Post-order Traversal

// OJ: https://leetcode.com/problems/increasing-order-search-tree/
// 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;
    }
};

Java

  • /**
     * Definition for a binary tree node.
     * public class TreeNode {
     *     int val;
     *     TreeNode left;
     *     TreeNode right;
     *     TreeNode(int x) { val = x; }
     * }
     */
    class Solution {
        public TreeNode increasingBST(TreeNode root) {
            if (root == null || root.left == null && root.right == null)
                return root;
            List<Integer> inorderTraversal = inorderTraversal(root);
            TreeNode newRoot = new TreeNode(inorderTraversal.get(0));
            TreeNode temp = newRoot;
            int size = inorderTraversal.size();
            for (int i = 1; i < size; i++) {
                TreeNode node = new TreeNode(inorderTraversal.get(i));
                temp.right = node;
                temp = temp.right;
            }
            return newRoot;
        }
    
        public List<Integer> inorderTraversal(TreeNode root) {
            List<Integer> inorderTraversal = new ArrayList<Integer>();
            Stack<TreeNode> stack = new Stack<TreeNode>();
            TreeNode node = root;
            while (!stack.isEmpty() || node != null) {
                while (node != null) {
                    stack.push(node);
                    node = node.left;
                }
                TreeNode visitNode = stack.pop();
                inorderTraversal.add(visitNode.val);
                node = visitNode.right;
            }
            return inorderTraversal;
        }
    }
    
  • // OJ: https://leetcode.com/problems/increasing-order-search-tree
    // Time: O(N)
    // Space: O(H)
    class Solution {
        TreeNode *prev = nullptr;
    public:
        TreeNode* increasingBST(TreeNode* root) {
            if (!root) return nullptr;
            auto head = increasingBST(root->left);
            root->left = nullptr;
            if (prev) prev->right = root;
            prev = root;
            root->right = increasingBST(root->right);
            return head ? head : root;
        }
    };
    
  • # Definition for a binary tree node.
    # class TreeNode(object):
    #     def __init__(self, x):
    #         self.val = x
    #         self.left = None
    #         self.right = None
    
    class Solution(object):
        def increasingBST(self, root):
            """
            :type root: TreeNode
            :rtype: TreeNode
            """
            array = self.inOrder(root)
            if not array:
                return None
            newRoot = TreeNode(array[0])
            curr = newRoot
            for i in range(1, len(array)):
                curr.right =TreeNode(array[i])
                curr = curr.right
            return newRoot
            
        def inOrder(self, root):
            if not root:
                return []
            array = []
            array.extend(self.inOrder(root.left))
            array.append(root.val)
            array.extend(self.inOrder(root.right))
            return array
    

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