[LintCode] Convert BST to Greater Tree-优快云博客

本文介绍如何将二叉搜索树（BST）转换为大于之树，即每个节点的值等于原值加上所有右侧节点值之和。提供两种实现方法：递归与迭代，并详细解释了算法原理。

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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

Given a binary search Tree ｀{5,2,13}｀:

              5
            /   \
           2     13

Return the root of new tree

             18
            /   \
          20     13

An in order traversal of a binary search tree gives an increasing sorted keys. For this problem, we need to visit all the keys in decreasing order. So we need to

use the in order traversal with a twist: right subtree -> root -> left subtree.

Solution 1. Recursion

 1 public class Solution {
 2     private int sum = 0; 
 3     public TreeNode convertBST(TreeNode root) {
 4         helper(root);
 5         return root;
 6     }
 7     private void helper(TreeNode node) {
 8         if(node == null) {
 9             return;
10         }
11         helper(node.right);
12         node.val += sum;
13         sum = node.val;
14         helper(node.left);
15     }
16 }



Solution 2. Iterative

 1 public class Solution {
 2     public TreeNode convertBST(TreeNode root) {
 3         Stack<TreeNode> stack = new Stack<TreeNode>();
 4         TreeNode curr = root;
 5         int sum = 0;
 6         
 7         while(curr != null || !stack.isEmpty()) {
 8             while(curr != null) {
 9                 stack.push(curr);
10                 curr = curr.right;
11             }
12             curr = stack.pop();
13             curr.val += sum;
14             sum = curr.val;
15             curr = curr.left;
16         }
17         return root;
18     }
19 }