一:题目

二:上码

/*** 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 {/**思路:1:确定递归函数的参数跟返回值2.确定终止条件当遍历为空的时候  我们需要  return   */public void  order(TreeNode root,List<Integer> list) {if (root == null) return ;order(root.left,list);list.add(root.val);order(root.right,list);}public int getSum(List<Integer> list,int nums) {int sum = 0;for (int i = 0; i < list.size(); i++) {if (list.get(i) >= nums) sum += list.get(i);}return sum;}public void getAns(TreeNode root,List<Integer>list) {if (root == null) return ;int sum = getSum(list,root.val);root.val = sum;getAns(root.left,list);getAns(root.right,list);}public TreeNode convertBST(TreeNode root) {List<Integer>list =  new ArrayList<>();order(root,list);getAns(root,list);return root;}
}