Given 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.

Example:

Input: The root of a Binary Search Tree like this:
              5
            /   \
           2     13

Output: The root of a Greater Tree like this:
             18
            /   \
          20     13

题目链接：https://leetcode.com/problems/convert-bst-to-greater-tree/

题目分析：

法一：最直接的方法，两次中序，第一次求后缀和，第二次赋值

9ms，时间击败15.57%，直接DFS 2ms，时间击败76.34%

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode convertBST(TreeNode root) {
        if (root == null || (root.left == null && root.right == null)) {
            return root;
        }
        TreeNode ans = root;
        List<Integer> values = new ArrayList<>();
        Stack<TreeNode> stk = new Stack<>();
        while (root != null || !stk.empty()) {
            while (root != null) {
                stk.push(root);
                root = root.left;
            }
            if (!stk.empty()) {
                root = stk.peek();
                stk.pop();
                values.add(root.val);
                root = root.right;
            }
        }
        int n = values.size();
        int[] suffixSum = new int[n];
        suffixSum[n - 1] = values.get(n - 1);
        for (int i = n - 2; i >= 0; i--) {
            suffixSum[i] = values.get(i) + suffixSum[i + 1];
        }
        stk.clear();
        int pos = 0;
        root = ans;
        while (root != null || !stk.empty()) {
            while (root != null) {
                stk.push(root);
                root = root.left;
            }
            if (!stk.empty()) {
                root = stk.peek();
                stk.pop();
                root.val = suffixSum[pos++];
                root = root.right;
            }
        }
        return ans;
    }
}

法二：按右中左的顺序做中序，直接累加结点值即可

0ms，时间击败100%

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    
    int suffixSum = 0;
    
    public void DFS(TreeNode root) {
        if (root == null) {
            return;
        }
        DFS(root.right);
        suffixSum += root.val;
        root.val = suffixSum;
        DFS(root.left);
    }
    
    public TreeNode convertBST(TreeNode root) {
        if (root == null || (root.left == null && root.right == null)) {
            return root;
        }
        DFS(root);
        return root;
    }
}