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; }
}
}
}