leetcode538

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    Stack<TreeNode> S = new Stack<TreeNode>();

        private void postNode(TreeNode node)
        {
            if (node != null)
            {
                if (node.left != null)
                {
                    postNode(node.left);
                }
                S.Push(node);
                if (node.right != null)
                {
                    postNode(node.right);
                }

            }
        }

        public TreeNode ConvertBST(TreeNode root)
        {            
            postNode(root);
            var list = S.ToList();
            var sum=0;

            foreach (var l in list)
            {
                sum += l.val;
                l.val = sum;
            }

            return root;
        }
}

https://leetcode.com/problems/convert-bst-to-greater-tree/#/description

补充一个python的实现，思路：使用中序遍历，在递归过程中累加节点和。

其实是把第一种方案的代码的两次循环合并成了一次，都是使用中序遍历的思想。

class Solution(object):
    def __init__(self):
        self.total = 0

    def convertBST(self, root):
        if root is not None:
            self.convertBST(root.right)
            self.total += root.val
            root.val = self.total
            self.convertBST(root.left)
        return root