LeetCode 538. Convert BST to Greater Tree

原题链接在这里：https://leetcode.com/problems/convert-bst-to-greater-tree/description/

题目：

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

题解：

反着inorder走, 右中左, 更新sum.

Time Complexity: O(n), node数.

Space: O(logn), stack space.

AC Java:

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

Iteration做法. 

Time Complexity: O(n). Space: O(logn).

AC Java: 

/**
 * 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) {
        int sum = 0;
        Stack<TreeNode> stk = new Stack<TreeNode>();
        TreeNode cur = root;
        
        while(!stk.isEmpty() || cur!=null){
            if(cur != null){
                stk.push(cur);
                cur = cur.right;
            }else{
                cur = stk.pop();
                sum += cur.val;
                cur.val = sum;
                cur = cur.left;
            }
        }
        return root;
    }
}

类似Binary Tree Inorder Traversal.