# 1305. All Elements in Two Binary Search Trees

## Description

Given two binary search trees root1 and root2, return a list containing all the integers from both trees sorted in ascending order.

Example 1:

Input: root1 = [2,1,4], root2 = [1,0,3]
Output: [0,1,1,2,3,4]


Example 2:

Input: root1 = [1,null,8], root2 = [8,1]
Output: [1,1,8,8]


Constraints:

• The number of nodes in each tree is in the range [0, 5000].
• -105 <= Node.val <= 105

## Solutions

• /**
* 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 List<Integer> getAllElements(TreeNode root1, TreeNode root2) {
List<Integer> t1 = new ArrayList<>();
List<Integer> t2 = new ArrayList<>();
dfs(root1, t1);
dfs(root2, t2);
return merge(t1, t2);
}

private void dfs(TreeNode root, List<Integer> t) {
if (root == null) {
return;
}
dfs(root.left, t);
dfs(root.right, t);
}

private List<Integer> merge(List<Integer> t1, List<Integer> t2) {
List<Integer> ans = new ArrayList<>();
int i = 0, j = 0;
while (i < t1.size() && j < t2.size()) {
if (t1.get(i) <= t2.get(j)) {
} else {
}
}
while (i < t1.size()) {
}
while (j < t2.size()) {
}
return ans;
}
}

• /**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode() : val(0), left(nullptr), right(nullptr) {}
*     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
*     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
vector<int> getAllElements(TreeNode* root1, TreeNode* root2) {
vector<int> t1;
vector<int> t2;
dfs(root1, t1);
dfs(root2, t2);
return merge(t1, t2);
}

void dfs(TreeNode* root, vector<int>& t) {
if (!root) return;
dfs(root->left, t);
t.push_back(root->val);
dfs(root->right, t);
}

vector<int> merge(vector<int>& t1, vector<int>& t2) {
vector<int> ans;
int i = 0, j = 0;
while (i < t1.size() && j < t2.size()) {
if (t1[i] <= t2[j])
ans.push_back(t1[i++]);
else
ans.push_back(t2[j++]);
}
while (i < t1.size()) ans.push_back(t1[i++]);
while (j < t2.size()) ans.push_back(t2[j++]);
return ans;
}
};

• # Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
def getAllElements(self, root1: TreeNode, root2: TreeNode) -> List[int]:
def dfs(root, t):
if root is None:
return
dfs(root.left, t)
t.append(root.val)
dfs(root.right, t)

def merge(t1, t2):
ans = []
i = j = 0
while i < len(t1) and j < len(t2):
if t1[i] <= t2[j]:
ans.append(t1[i])
i += 1
else:
ans.append(t2[j])
j += 1
while i < len(t1):
ans.append(t1[i])
i += 1
while j < len(t2):
ans.append(t2[j])
j += 1
return ans

t1, t2 = [], []
dfs(root1, t1)
dfs(root2, t2)
return merge(t1, t2)


• /**
* Definition for a binary tree node.
* type TreeNode struct {
*     Val int
*     Left *TreeNode
*     Right *TreeNode
* }
*/
func getAllElements(root1 *TreeNode, root2 *TreeNode) []int {
var dfs func(root *TreeNode) []int
dfs = func(root *TreeNode) []int {
if root == nil {
return []int{}
}
left := dfs(root.Left)
right := dfs(root.Right)
left = append(left, root.Val)
left = append(left, right...)
return left
}
merge := func(t1, t2 []int) []int {
var ans []int
i, j := 0, 0
for i < len(t1) && j < len(t2) {
if t1[i] <= t2[j] {
ans = append(ans, t1[i])
i++
} else {
ans = append(ans, t2[j])
j++
}
}
for i < len(t1) {
ans = append(ans, t1[i])
i++
}
for j < len(t2) {
ans = append(ans, t2[j])
j++
}
return ans
}
t1, t2 := dfs(root1), dfs(root2)
return merge(t1, t2)
}

• /**
* Definition for a binary tree node.
* class TreeNode {
*     val: number
*     left: TreeNode | null
*     right: TreeNode | null
*     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
*         this.val = (val===undefined ? 0 : val)
*         this.left = (left===undefined ? null : left)
*         this.right = (right===undefined ? null : right)
*     }
* }
*/

function getAllElements(root1: TreeNode | null, root2: TreeNode | null): number[] {
const res = [];
const stacks = [[], []];
while (root1 != null || stacks[0].length !== 0 || root2 != null || stacks[1].length !== 0) {
if (root1 != null) {
stacks[0].push(root1);
root1 = root1.left;
} else if (root2 != null) {
stacks[1].push(root2);
root2 = root2.left;
} else {
if (
(stacks[0][stacks[0].length - 1] ?? { val: Infinity }).val <
(stacks[1][stacks[1].length - 1] ?? { val: Infinity }).val
) {
const { val, right } = stacks[0].pop();
res.push(val);
root1 = right;
} else {
const { val, right } = stacks[1].pop();
res.push(val);
root2 = right;
}
}
}
return res;
}


• // 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
//     }
//   }
// }
use std::cell::RefCell;
use std::rc::Rc;
impl Solution {
pub fn get_all_elements(
root1: Option<Rc<RefCell<TreeNode>>>,
root2: Option<Rc<RefCell<TreeNode>>>
) -> Vec<i32> {
fn dfs(root: &Option<Rc<RefCell<TreeNode>>>, t: &mut Vec<i32>) {
if let Some(root) = root {
dfs(&root.borrow().left, t);
t.push(root.borrow().val);
dfs(&root.borrow().right, t);
}
}

let mut t1 = Vec::new();
let mut t2 = Vec::new();
dfs(&root1, &mut t1);
dfs(&root2, &mut t2);

let mut ans = Vec::new();
let mut i = 0;
let mut j = 0;
while i < t1.len() && j < t2.len() {
if t1[i] < t2[j] {
ans.push(t1[i]);
i += 1;
} else {
ans.push(t2[j]);
j += 1;
}
}
while i < t1.len() {
ans.push(t1[i]);
i += 1;
}
while j < t2.len() {
ans.push(t2[j]);
j += 1;
}
ans
}
}