package leetcode_1To300;

import java.util.Stack;

/**
 * 本代码来自 Cspiration，由 @Cspiration 提供
 * 题目来源：http://leetcode.com
 * - Cspiration 致力于在 CS 领域内帮助中国人找到工作，让更多海外国人受益
 * - 现有课程：Leetcode Java 版本视频讲解（1-900题）（上）（中）（下）三部
 * - 算法基础知识（上）（下）两部；题型技巧讲解（上）（下）两部
 * - 节省刷题时间，效率提高2-3倍，初学者轻松一天10题，入门者轻松一天20题
 * - 讲师：Edward Shi
 * - 官方网站：https://cspiration.com
 * - 版权所有，转发请注明出处
 */
public class _99_RecoverBinarySearchTree {
    /**
     * 99. Recover Binary Search Tree
     * Two elements of a binary search tree (BST) are swapped by mistake.

     Recover the tree without changing its structure.
          6
         / \
        8  1
      /  \
     0   3
        / \
       2  5

     time : O(n)
     space : O(n)


     */

    TreeNode first = null;
    TreeNode second = null;
    TreeNode prev = null;

    public void recoverTree(TreeNode root) {
        if (root == null) return;
        helper(root);
        int temp = first.val;
        first.val = second.val;
        second.val = temp;
    }
    public void helper(TreeNode root) {
        if (root == null) return;
        helper(root.left);
        if (prev != null && prev.val >= root.val) {
            if (first == null) first = prev;
            second = root;
        }
        prev = root;
        helper(root.right);
    }

    public void recoverTree2(TreeNode root) {
        if (root == null) return;
        TreeNode first = null;
        TreeNode second = null;
        TreeNode prev = null;

        TreeNode cur = root;
        Stack<TreeNode> stack = new Stack<>();
        while (!stack.isEmpty() || cur != null) {
            if (cur != null) {
                stack.push(cur);
                cur = cur.left;
            } else {
                cur = stack.pop();
                if (prev != null && cur.val <= prev.val) {
                    if (first == null) first = prev;
                    second = cur;
                }
                prev = cur;
                cur = cur.right;
            }
        }
        int temp = first.val;
        first.val = second.val;
        second.val = temp;
    }
}