利用好BST的特性,解起來並不難node
1 /** 2 * Definition for a binary tree node. 3 * public class TreeNode { 4 * int val; 5 * TreeNode left; 6 * TreeNode right; 7 * TreeNode(int x) { val = x; } 8 * } 9 */ 10 public class Solution { 11 private int sum = 0; 12 public TreeNode convertBST(TreeNode root) { 13 if (root == null) return null; 14 convertBST(root.right); 15 root.val += sum; 16 sum = root.val; 17 convertBST(root.left); 18 return root; 19 } 20 }