In this post, we are going to solve the 1305. All Elements in Two Binary Search Trees problem of Leetcode. This problem 1305. All Elements in Two Binary Search Trees is a Leetcode medium level problem. Let’s see the code, 1305. All Elements in Two Binary Search Trees – Leetcode Solution.

Problem

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

Now, let’s see the code of 1305. All Elements in Two Binary Search Trees – Leetcode Solution.

All Elements in Two Binary Search Trees – Leetcode Solution

1305. All Elements in Two Binary Search Trees – Solution in Java

/**
 * 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> list1 = new ArrayList<Integer>();
        List<Integer> list2 = new ArrayList<Integer>();
        List<Integer> ans = new ArrayList<Integer>();
        
        inorder(root1,list1);
        inorder(root2,list2);
        
        int i=0,j=0;
        
        while(i<list1.size() && j<list2.size()){
            if(list1.get(i) < list2.get(j)){
                ans.add(list1.get(i));
                i++;
            }
            else{
                 ans.add(list2.get(j));
                j++;
            }
        }
        
        while(i<list1.size()){
            ans.add(list1.get(i));
            i++;
        }
         while(j<list2.size()){
            ans.add(list2.get(j));
            j++;
        }
        
        return ans;
        
    }
    
    public void inorder(TreeNode root, List<Integer> list){
        if(root == null)return;
        inorder(root.left,list);
        list.add(root.val);
        inorder(root.right,list);
    }
}

1305. All Elements in Two Binary Search Trees – Solution in C++

/**
 * 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) {
        stack<TreeNode *> st1, st2;
        vector<int> res;
        
        while(root1 || root2 || !st1.empty() || !st2.empty()){
            while(root1){
                st1.push(root1);
                root1 = root1->left;
            }
            while(root2){
                st2.push(root2);
                root2 = root2->left;
            }
            if(st2.empty() || (!st1.empty() && st1.top()->val <= st2.top()->val)){
                root1 = st1.top();
                st1.pop();
                res.push_back(root1->val);
                root1 = root1->right;
            }
            else{
                root2 = st2.top();
                st2.pop();
                res.push_back(root2->val);
                root2 = root2->right;
            }
        }
        return res;
    }
};

1305. All Elements in Two Binary Search Trees– Solution in Python

# 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:
    def getAllElements(self, root1, root2):
        def inorder(root, lst):
            if not root: return
            inorder(root.left, lst)
            lst.append(root.val)
            inorder(root.right, lst)
        
        lst1, lst2 = [], []
        inorder(root1, lst1)
        inorder(root2, lst2)
        
        i1, i2, res = 0, 0, []
        s1, s2 = len(lst1), len(lst2)
        
        while i1 < s1 and i2 < s2:
            if lst1[i1] < lst2[i2]:
                res += [lst1[i1]]
                i1 += 1
            else:
                res += [lst2[i2]]
                i2 += 1
                
        return res + lst1[i1:] + lst2[i2:]
        

Note: This problem 1305. All Elements in Two Binary Search Trees is generated by Leetcode but the solution is provided by CodingBroz. This tutorial is only for Educational and Learning purpose.