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Formatted question description: https://leetcode.ca/all/1305.html

# 1305. All Elements in Two Binary Search Trees (Medium)

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 = [0,-10,10], root2 = [5,1,7,0,2]
Output: [-10,0,0,1,2,5,7,10]


Example 3:

Input: root1 = [], root2 = [5,1,7,0,2]
Output: [0,1,2,5,7]


Example 4:

Input: root1 = [0,-10,10], root2 = []
Output: [-10,0,10]


Example 5:

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


Constraints:

• Each tree has at most 5000 nodes.
• Each node's value is between [-10^5, 10^5].

Related Topics:
Sort, Tree

## Solution 1.

• /**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
class Solution {
public List<Integer> getAllElements(TreeNode root1, TreeNode root2) {
List<Integer> elements1 = inorderTraversal(root1);
List<Integer> elements2 = inorderTraversal(root2);
List<Integer> allElements = new ArrayList<Integer>();
int size1 = elements1.size(), size2 = elements2.size();
int index1 = 0, index2 = 0;
while (index1 < size1 && index2 < size2) {
int element1 = elements1.get(index1), element2 = elements2.get(index2);
if (element1 <= element2) {
index1++;
} else {
index2++;
}
}
while (index1 < size1) {
int element1 = elements1.get(index1);
index1++;
}
while (index2 < size2) {
int element2 = elements2.get(index2);
index2++;
}
return allElements;
}

public List<Integer> inorderTraversal(TreeNode root) {
List<Integer> inorderTraversal = new ArrayList<Integer>();
Stack<TreeNode> stack = new Stack<TreeNode>();
TreeNode node = root;
while (!stack.isEmpty() || node != null) {
while (node != null) {
stack.push(node);
node = node.left;
}
TreeNode visitNode = stack.pop();
node = visitNode.right;
}
return inorderTraversal;
}
}

############

/**
* 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;
}
}

• // OJ: https://leetcode.com/problems/all-elements-in-two-binary-search-trees/
// Time: O(A + B)
// Space: O(HA + HB)
class BstIterator {
stack<TreeNode*> s;
void pushNodes(TreeNode *node) {
for (; node; node = node->left) s.push(node);
}
public:
BstIterator(TreeNode *root) {
pushNodes(root);
}
int peek() {
return hasNext() ? s.top()->val : INT_MIN;
}
bool hasNext() {
return s.size();
}
void next() {
if (!hasNext()) return;
auto n = s.top();
s.pop();
pushNodes(n->right);
}
};
class Solution {
public:
vector<int> getAllElements(TreeNode* a, TreeNode* b) {
BstIterator i(a), j(b);
vector<int> ans;
while (i.hasNext() || j.hasNext()) {
if (!j.hasNext() || (i.hasNext() && i.peek() <= j.peek())) {
ans.push_back(i.peek());
i.next();
} else {
ans.push_back(j.peek());
j.next();
}
}
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.
# class TreeNode(object):
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution(object):
def getAllElements(self, root1, root2):
"""
:type root1: TreeNode
:type root2: TreeNode
:rtype: List[int]
"""
nums1 = []
nums2 = []
self.inOrder(root1, nums1)
self.inOrder(root2, nums2)
return self.merge(nums1, nums2)

def inOrder(self, root, nums):
if not root:
return
self.inOrder(root.left, nums)
nums.append(root.val)
self.inOrder(root.right, nums)

def merge(self, nums1, nums2):
res = []
p1 = 0
p2 = 0
while p1 < len(nums1) and p2 < len(nums2):
if nums1[p1] < nums2[p2]:
res.append(nums1[p1])
p1 += 1
else:
res.append(nums2[p2])
p2 += 1
if p1 < len(nums1):
res.extend(nums1[p1:])
if p2 < len(nums2):
res.extend(nums2[p2:])
return res

• /**
* 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
}
}