Leetcode 538 - Convert BST to Greater Tree

Link to problem

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 the 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.

Solution

From the BST’s 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.

• We know that the current node’s new value will be sum of left subtree + current node’s value.
• The sum of left subtree + the current node needs to be passed down to the right subtree. A leaf node’s sum will be the sum passed down from it’s parent + the value of itself.
• The left most node of the a subtree will contain the sum of the whole subtree, which needs to be returned to process other subproblems.

And there, we could write a recursion from the bullet points above.

Implementation

/**
 * 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:

    int runner(TreeNode* root, int sum) {
        if (!root) return sum;

        int right = runner(root->right, sum);
        root->val += right;
        int left = runner(root->left, root->val);

        return left;
    }

    TreeNode* convertBST(TreeNode* root) {

       runner(root, 0);

       return root;
    }
};