1038. Binary Search Tree to Greater Sum Tree

QUESTION:

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:

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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.

EXPLANATION:

这道题目就很简单了，其实就是先序遍历的变种，先序都是从左边开始，但是这次需要从右边开始。 这样遍历结束就是按照需要的顺序进行的，再设置一个sum来计算总和，就可以解决了。

SOLUTION:

class Solution {
    public TreeNode bstToGst(TreeNode root) {
        bstToGstHelper(root);
        return root;
    }
    public int bstToGstSum = 0;
    public void bstToGstHelper(TreeNode root) {
        if(root == null) return;
        if(root.right!=null) bstToGstHelper(root.right);
        bstToGstSum+=root.val;
        root.val = bstToGstSum;
        if(root.left!=null) bstToGstHelper(root.left);
    }
}
--> return <--