Formatted question description: https://leetcode.ca/all/1038.html

1038. Binary Search Tree to Greater Sum Tree (Medium)

Given the root of a binary search tree with distinct values, modify it so that every node has a new value equal to the sum of the values of the original tree that are greater than or equal to node.val.

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.

Example 1:

Input: [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]

Note:

The number of nodes in the tree is between 1 and 100.
Each node will have value between 0 and 100.
The given tree is a binary search tree.

Companies:
Amazon

Related Topics:
Binary Search Tree

Solution 1.

// OJ: https://leetcode.com/problems/binary-search-tree-to-greater-sum-tree/
// Time: O(N)
// Space: O(N)
class Solution {
private:
    int sum = 0;
public:
    TreeNode* bstToGst(TreeNode* root) {
        if (!root) return NULL;
        bstToGst(root->right);
        root->val = (sum += root->val);
        bstToGst(root->left);
        return root;
    }
};

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

############

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode bstToGst(TreeNode root) {
        int s = 0;
        TreeNode node = root;
        while (root != null) {
            if (root.right == null) {
                s += root.val;
                root.val = s;
                root = root.left;
            } else {
                TreeNode next = root.right;
                while (next.left != null && next.left != root) {
                    next = next.left;
                }
                if (next.left == null) {
                    next.left = root;
                    root = root.right;
                } else {
                    s += root.val;
                    root.val = s;
                    next.left = null;
                    root = root.left;
                }
            }
        }
        return node;
    }
}

// OJ: https://leetcode.com/problems/binary-search-tree-to-greater-sum-tree/
// Time: O(N)
// Space: O(H)
class Solution {
    int sum = 0;
public:
    TreeNode* bstToGst(TreeNode* root) {
        if (!root) return nullptr;
        bstToGst(root->right);
        root->val = (sum += root->val);
        bstToGst(root->left);
        return root;
    }
};

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def bstToGst(self, root: TreeNode) -> TreeNode:
        s = 0
        node = root
        while root:
            if root.right is None:
                s += root.val
                root.val = s
                root = root.left
            else:
                next = root.right
                while next.left and next.left != root:
                    next = next.left
                if next.left is None:
                    next.left = root
                    root = root.right
                else:
                    s += root.val
                    root.val = s
                    next.left = None
                    root = root.left
        return node

############

# 1038. Binary Search Tree to Greater Sum Tree
# https://leetcode.com/problems/binary-search-tree-to-greater-sum-tree/

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    curr = 0
    def bstToGst(self, root: TreeNode) -> TreeNode:
        if root.right: self.bstToGst(root.right)
        root.val = self.curr = self.curr + root.val
        if root.left: self.bstToGst(root.left)
        
        return root

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func bstToGst(root *TreeNode) *TreeNode {
	s := 0
	node := root
	for root != nil {
		if root.Right == nil {
			s += root.Val
			root.Val = s
			root = root.Left
		} else {
			next := root.Right
			for next.Left != nil && next.Left != root {
				next = next.Left
			}
			if next.Left == nil {
				next.Left = root
				root = root.Right
			} else {
				s += root.Val
				root.Val = s
				next.Left = nil
				root = root.Left
			}
		}
	}
	return node
}

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function bstToGst(root: TreeNode | null): TreeNode | null {
    let cur = root;
    let sum = 0;
    while (cur != null) {
        const { val, left, right } = cur;
        if (right == null) {
            sum += val;
            cur.val = sum;
            cur = left;
        } else {
            let next = right;
            while (next.left != null && next.left != cur) {
                next = next.left;
            }
            if (next.left == null) {
                next.left = cur;
                cur = right;
            } else {
                next.left = null;
                sum += val;
                cur.val = sum;
                cur = left;
            }
        }
    }
    return root;
}

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {TreeNode}
 */
var bstToGst = function (root) {
    let s = 0;
    function dfs(root) {
        if (!root) {
            return;
        }
        dfs(root.right);
        s += root.val;
        root.val = s;
        dfs(root.left);
    }
    dfs(root);
    return root;
};

// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::rc::Rc;
use std::cell::RefCell;
impl Solution {
    fn dfs(root: &mut Option<Rc<RefCell<TreeNode>>>, mut sum: i32) -> i32 {
        if let Some(node) = root {
            let mut node = node.as_ref().borrow_mut();
            sum = Self::dfs(&mut node.right, sum) + node.val;
            node.val = sum;
            sum = Self::dfs(&mut node.left, sum);
        }
        sum
    }

    pub fn bst_to_gst(mut root: Option<Rc<RefCell<TreeNode>>>) -> Option<Rc<RefCell<TreeNode>>> {
        Self::dfs(&mut root, 0);
        root
    }
}

1038 - Binary Search Tree to Greater Sum Tree

1038. Binary Search Tree to Greater Sum Tree (Medium)

Solution 1.

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1038. Binary Search Tree to Greater Sum Tree (Medium)

Solution 1.

All Problems

All Solutions