leetcode

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     13Output: The root of a Greater Tree like this:             18            /   \          20     13

题意

给定一个二叉查找树，将转化为更大的树，在这棵大树中，每个原来结点的键值都变为 （原来的键值 + 所有比该键原来键值大的键之和）。

分析及解答

解法1：(从大到小，利用队列，累积)

• 【处理的顺序：从小到大】根据题意，确定积累的规则，和处理的顺序，从大到小处理，这样做的好处是后面的小数能够利用前面计算累积的结果（sum）,同时确保小数只看到比其大的数即可，因为其只依赖于比其大的数。

public class Solution {public TreeNode convertBST(TreeNode root) {if (root == null) {return null;}Queue<TreeNode> queue = new LinkedList<>();initQueue(root, queue);TreeNode pre = queue.poll();TreeNode current = queue.poll();while(current != null){current.val = current.val + pre.val;            pre = current;current = queue.poll();}return root;}public void initQueue(TreeNode root, Queue<TreeNode> queue) {if (root == null) {return;}initQueue(root.right, queue);queue.add(root);initQueue(root.left,queue);}}

解法2：（思想同上，遍历 + 全局变量）

public class Solution {    int sum = 0;        public TreeNode convertBST(TreeNode root) {        convert(root);        return root;    }        public void convert(TreeNode cur) {        if (cur == null) return;        convert(cur.right);        cur.val += sum;        sum = cur.val;        convert(cur.left);    }}