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.
Note: This question is the same as 1038: https://leetcode.com/problems/binary-search-tree-to-greater-sum-tree/
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]
Constraints:
- The number of nodes in the tree is in the range
[0, 104]. -104 <= Node.val <= 104- All the values in the tree are unique.
rootis guaranteed to be a valid binary search tree.
struct Solution;
use rustgym_util::*;
trait Inorder {
fn inorder(&mut self, sum: &mut i32);
}
impl Inorder for TreeLink {
fn inorder(&mut self, sum: &mut i32) {
if let Some(node) = self {
let mut node = node.borrow_mut();
node.right.inorder(sum);
*sum += node.val;
node.val = *sum;
node.left.inorder(sum);
}
}
}
impl Solution {
fn convert_bst(mut root: TreeLink) -> TreeLink {
let mut sum = 0;
root.inorder(&mut sum);
root
}
}
#[test]
fn test() {
let root = tree!(5, tree!(2), tree!(13));
let greater = tree!(18, tree!(20), tree!(13));
assert_eq!(Solution::convert_bst(root), greater);
}