538. Convert BST to Greater Tree

Given 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.

Example:

Input: The root of a Binary Search Tree like this:
              5
            /   \
           2     13

Output: The root of a Greater Tree like this:
             18
            /   \
          20     13

Solution:

Method 1: Recursive

The in-order traversal of BST is in ascending order.

If we reverse the in-order traversal with right-root-left order, the result will be in depending order.

Therefore, we can use a reversed in-order traversal, and update each node's value by adding its previous node's value.



Code:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public int sum = 0;
    public TreeNode convertBST(TreeNode root) {
        if (root == null) {
            return null;
        }
        convertBST(root.right);
        sum += root.val;
        root.val = sum;
        convertBST(root.left);
        return root;
    }
}

Method 2: Iterative


Code:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode convertBST(TreeNode root) {
        if (root == null) {
            return null;
        }
        Stack<TreeNode> stack = new Stack<>();
        int sum = 0;
        TreeNode curr = root;
        while (curr != null || !stack.isEmpty()) {
            while (curr != null) {
                stack.push(curr);
                curr = curr.right;
            }
            curr = stack.pop();
            sum += curr.val;
            curr.val = sum;
            curr = curr.left;
        }
        return root;
    }
}