538. Convert BST to Greater Tree • Algorithm Solutions

1. Description

Given the root of a Binary Search Tree (BST), convert it to a Greater Tree such that every key of the original BST is changed to the original key plus sum of all keys greater than the original key in BST.
As a reminder, a binary search tree is a tree that satisfies these constraints:

The left subtree of a node contains only nodes with keys less than the node’s key.
The right subtree of a node contains only nodes with keys greater than the node’s key.
Both the left and right subtrees must also be binary search trees.

2. Note

This question is the same as 1038: https://leetcode.com/problems/binary-search-tree-to-greater-sum-tree/

3. Example

Example 1:
Input: root = [4,1,6,0,2,5,7,null,null,null,3,null,null,null,8]
Output: [30,36,21,36,35,26,15,null,null,null,33,null,null,null,8]

Example 2:
Input: root = [0,null,1]
Output: [1,null,1]

Example 3:
Input: root = [1,0,2]
Output: [3,3,2]

Example 4:
Input: root = [3,2,4,1]
Output: [7,9,4,10]

4. Constraints

The number of nodes in the tree is in the range [0, $10^4$].
$-10^4$ <= Node.val <= $10^4$
All the values in the tree are unique.
root is guaranteed to be a valid binary search tree.

5. Solutions

My Accepted Solution

n is the number of nodes in m_root
Time complexity: O(n)
Space complexity: O(n)

class Solution 
{
private:
    vector<TreeNode *> nodesInPreorder;

    void preorderTraverse(TreeNode *i_root)
    {
        if(i_root->left) preorderTraverse(i_root->left);
        
        nodesInPreorder.push_back(i_root);
        
        if(i_root->right) preorderTraverse(i_root->right);        
    } 
    
public:
    // TreeNode* convertBST(TreeNode* root)
    TreeNode* convertBST(TreeNode *m_root) 
    {
        if(!m_root) return nullptr;
        
        preorderTraverse(m_root);
        for(int i = nodesInPreorder.size() - 2; i >= 0; i--)
        {
            nodesInPreorder[i]->val += nodesInPreorder[i+1]->val;
        }
        
        return m_root;
    }
};

5.1 Reverse Preorder Traversal && Iteration

n is the number of nodes in m_root
Time complexity: O(n)
Space complexity: O(n)

class Solution 
{
public:
    // TreeNode* convertBST(TreeNode* root)
    TreeNode* convertBST(TreeNode *m_root) 
    {
        int postSum = 0;
        auto iter = m_root;
        stack<TreeNode *> parentNodes;
        
        while(iter || !parentNodes.empty())
        {
            // find the most right node
            while(iter)
            {
                parentNodes.push(iter);
                iter = iter->right;
            }
            
            iter = parentNodes.top();
            parentNodes.pop();
            
            postSum += iter->val;
            iter->val = postSum;
            
            iter = iter->left;
        }
        
        return m_root;
    }
};

5.2 Reverse Morris && Iteration

n is the number of nodes in m_root
Time complexity: O(n)
Space complexity: O(1)

class Solution 
{
public:
    // TreeNode* convertBST(TreeNode* root)
    TreeNode* convertBST(TreeNode *m_root) 
    {
        int postSum = 0;
        TreeNode *currentNode = m_root, *previousNode = nullptr;
        while(currentNode)
        {
            if(!currentNode->right)
            {
                postSum += currentNode->val;
                currentNode->val = postSum;
                
                currentNode = currentNode->left;
            }
            else
            {
                previousNode = currentNode->right;
                
                while(previousNode->left && previousNode->left != currentNode)
                    previousNode = previousNode->left;
                
                if(!previousNode->left)
                {
                    previousNode->left = currentNode;
                    currentNode = currentNode->right;
                }
                else
                {
                    previousNode->left = nullptr;
                    
                    postSum += currentNode->val;
                    currentNode->val = postSum;
                    
                    currentNode = currentNode->left;
                }
            }
        }
        
        return m_root;
    }
};