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:

  1. The number of nodes in the tree is between 1 and 100.
  2. Each node will have value between 0 and 100.
  3. 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;
    }
};

Java

/**
 * 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;
    }
}

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