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538. Convert BST to Greater Tree

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.

 

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]

 

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.
  • root is guaranteed to be a valid binary search tree.

 

Note: This question is the same as 1038: https://leetcode.com/problems/binary-search-tree-to-greater-sum-tree/

Solutions

Solution 1

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# 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 convertBST(self, root: TreeNode) -> TreeNode:
        def dfs(root):
            nonlocal s
            if root is None:
                return
            dfs(root.right)
            s += root.val
            root.val = s
            dfs(root.left)

        s = 0
        dfs(root)
        return root

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/**
 * 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 {
    private int s;

    public TreeNode convertBST(TreeNode root) {
        dfs(root);
        return root;
    }

    private void dfs(TreeNode root) {
        if (root == null) {
            return;
        }
        dfs(root.right);
        s += root.val;
        root.val = s;
        dfs(root.left);
    }
}

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/**
 * 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 s = 0;

    TreeNode* convertBST(TreeNode* root) {
        dfs(root);
        return root;
    }

    void dfs(TreeNode* root) {
        if (!root) return;
        dfs(root->right);
        s += root->val;
        root->val = s;
        dfs(root->left);
    }
};

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/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func convertBST(root *TreeNode) *TreeNode {
    s := 0
    var dfs func(*TreeNode)
    dfs = func(root *TreeNode) {
        if root == nil {
            return
        }
        dfs(root.Right)
        s += root.Val
        root.Val = s
        dfs(root.Left)
    }
    dfs(root)
    return root
}

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/**
 * 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 convertBST = 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;
};

Solution 2