1038. Binary Search Tree to Greater Sum Tree

1038. Binary Search Tree to Greater Sum Tree

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 538: https://leetcode.com/problems/convert-bst-to-greater-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 [1, 100].
• 0 <= Node.val <= 100
• All the values in the tree are unique.
• root is guaranteed to be a valid binary search tree.

class Solution {
    int pre = 0;
    public TreeNode bstToGst(TreeNode root) {
        if(root == null) return null;
        if(root.right != null) bstToGst(root.right);
        pre = root.val + pre;
        root.val = pre;
        if(root.left != null) bstToGst(root.left);
        return root;
    }
}

因为要小的变成所有比他大的root之和，所以需要reverse inorder，右根左。根的时候更新pre和当前root的val，最后返回root。