use std::borrow::Borrow;
use std::cell::RefCell;
use std::collections::HashMap;
use std::ops::Index;
use std::rc::Rc;

/**
653. Two Sum IV - Input is a BST
https://leetcode.com/problems/two-sum-iv-input-is-a-bst/
Given the root of a Binary Search Tree and a target number k,
return true if there exist two elements in the BST such that their sum is equal to the given target.
*/
// Definition for a binary tree node.
#[derive(Debug, PartialEq, Eq)]
pub struct TreeNode {
    pub val: i32,
    pub left: Option<Rc<RefCell<TreeNode>>>,
    pub right: Option<Rc<RefCell<TreeNode>>>,
}

//
impl TreeNode {
    #[inline]
    pub fn new(val: i32) -> Self {
        TreeNode {
            val,
            left: None,
            right: None,
        }
    }
}

pub struct Solution {}

/*
solution: inorder + Hashmap, Time:O(n), Space:O(n)
*/
impl Solution {
    pub fn find_target(root: Option<Rc<RefCell<TreeNode>>>, k: i32) -> bool {
        let mut list: Vec<i32> = Vec::new();
        Self::in_order(root, &mut list);
        let mut map: HashMap<i32, i32> = HashMap::new();
        let mut i: i32 = 0;
        for item in list {
            let needToFind = k - item;
            if (map.contains_key(&needToFind)) {
                return true;
            }
            map.insert(item, i);
            i = i + 1;
        }
        return false;
    }

    fn in_order(root: Option<Rc<RefCell<TreeNode>>>, list: &mut Vec<i32>) {
        match root {
            Some(node) => {
                Self::in_order(node.as_ref().borrow().left.clone(), list);
                list.push(node.as_ref().borrow().val);
                Self::in_order(node.as_ref().borrow().right.clone(), list);
            }
            None => { return; }
        }
    }
}