538. Convert BST to Greater Tree

Leetcode Tree

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

分析

首先想到的办法是直接根据题目的描述一步一步做。第一步，遍历整个二叉树得到节点。第二步，计算每个节点的值。第三步，遍历二叉树给节点赋值。

private int position;
private List<Integer> list;

public TreeNode convertBST(TreeNode root) {
    list = new ArrayList<>();
    position = 0;
    inorderTraversal(root);
    for (int i = list.size() - 2; i >= 0; i--) {
        list.set(i, list.get(i) + list.get(i+1));
    }
    System.out.println(list);
    inorderTraversalSet(root);
    return root;

}

private void inorderTraversalSet(TreeNode root) {
    if (root == null) return;
    inorderTraversalSet(root.left);
    root.val = list.get(position++);
    inorderTraversalSet(root.right);
}

private void inorderTraversal(TreeNode root) {
    if (root == null) return;
    inorderTraversal(root.left);
    list.add(root.val);
    inorderTraversal(root.right);
}

一般的中序遍历的结果是依次递增的序列，那么可不可以变成依次递减的序列，这样的话，就可以通过递归来一次性赋值了。答案是可以：通过交换访问左子节点和右子节点的顺序。

private int sum = 0;
public TreeNode convertBST(TreeNode root) {
    if (root == null) return null;
    convertBST(root.right);
    sum += root.val;
    root.val = sum;
    convertBST(root.left);
    return root;
}