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:

 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 class Solution {
11     
12     int sum = 0;
13     
14     public TreeNode convertBST(TreeNode root) {
15         reverseInorder(root);
16         return root;
17     }
18     
19     private void reverseInorder(TreeNode root){
20         if(root == null){
21             return;
22         }
23         
24         reverseInorder(root.right);
25         root.val += sum;
26         sum = root.val;
27         reverseInorder(root.left);
28     }
29 }

Iteration做法. 

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

AC Java: 

 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 class Solution {
11     public TreeNode convertBST(TreeNode root) {
12         int sum = 0;
13         Stack<TreeNode> stk = new Stack<TreeNode>();
14         TreeNode cur = root;
15         
16         while(!stk.isEmpty() || cur!=null){
17             if(cur != null){
18                 stk.push(cur);
19                 cur = cur.right;
20             }else{
21                 cur = stk.pop();
22                 sum += cur.val;
23                 cur.val = sum;
24                 cur = cur.left;
25             }
26         }
27         return root;
28     }
29 }

类似Binary Tree Inorder Traversal.

posted @ 2017-09-13 13:21  Dylan_Java_NYC  阅读(125)  评论(0)  编辑  收藏  举报