Leetcode 538. Convert BST to Greater Tree

题目链接：https://leetcode.cn/problems/convert-bst-to-greater-tree/

方法一 迭代

1 方法思想

2 代码实现

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
        public TreeNode convertBST(TreeNode root) {
        if (root == null) return root;
        Stack<TreeNode> stack = new Stack<>();
        TreeNode cur = root;
        int sum = 0;
        while (!stack.isEmpty() || cur != null) {
            while (cur != null) {
                stack.push(cur);
                cur = cur.right;
            }

            int size = stack.size();
            cur = stack.pop();

            sum += cur.val;
            cur.val = sum;
            cur = cur.left;
        }
        return root;
    }
}

3 复杂度分析

时间复杂度：
空间复杂度：

4 涉及到知识点

5 总结

方法二 递归

1 方法思想

2 代码实现

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

3 复杂度分析

时间复杂度：
空间复杂度：

4 涉及到知识点

5 总结