All Elements in Two Binary Search Trees

• Create an empty list list.
• Define a recursive helper function collectElements that takes a TreeNode and the list. If the node is not null, it adds the node's value to the list and recursively calls itself for the left and right children.
• Call collectElements for root1 to add all its elements to list.
• Call collectElements for root2 to add all its elements to list.
• Use a standard library function (e.g., Collections.sort()) to sort list.
• Return the sorted list.

The core idea is to treat the trees as simple collections of numbers, ignoring their BST structure initially.

1. Initialization: We start by creating a dynamic array or list, let's call it elements, to hold all the integers from both trees.
2. Data Collection: We perform a traversal on the first tree, root1. For every node we visit, we add its value to the elements list. We repeat the same process for the second tree, root2. After this step, elements contains all values from both trees, but in an unsorted order.
3. Sorting: We apply a standard sorting algorithm to the elements list. Most programming languages provide a built-in sort function that is typically implemented with an efficient algorithm like Timsort or Quicksort.
4. Return: The now-sorted elements list is the final result.

1/**2 * Definition for a binary tree node.3 * public class TreeNode {4 *     int val;5 *     TreeNode left;6 *     TreeNode right;7 *     TreeNode() {}8 *     TreeNode(int val) { this.val = val; }9 *     TreeNode(int val, TreeNode left, TreeNode right) {10 *         this.val = val;11 *         this.left = left;12 *         this.right = right;13 *     }14 * }15 */16class Solution {17    public List<Integer> getAllElements(TreeNode root1, TreeNode root2) {18        List<Integer> list = new ArrayList<>();19        // Step 1 & 2: Traverse both trees and collect all elements20        collectElements(root1, list);21        collectElements(root2, list);22        23        // Step 3: Sort the combined list24        Collections.sort(list);25        26        // Step 4: Return the sorted list27        return list;28    }29    30    private void collectElements(TreeNode node, List<Integer> list) {31        if (node == null) {32            return;33        }34        list.add(node.val);35        collectElements(node.left, list);36        collectElements(node.right, list);37    }38}

• Create two empty lists, list1 and list2.
• Perform an in-order traversal on root1 and store the resulting sorted elements in list1.
• Perform an in-order traversal on root2 and store the resulting sorted elements in list2.
• Create an empty result list result.
• Use a two-pointer approach to merge the two sorted lists (list1 and list2) into the result list.
• Return result.

This method is more efficient because it takes advantage of the BST property.

1. In-order Traversal: We define a recursive inorder helper function. This function traverses a tree by first visiting the left subtree, then the node itself, and finally the right subtree. We call this function on root1 to populate a list, list1, and on root2 to populate another list, list2. Because of the in-order traversal on BSTs, both list1 and list2 will be sorted.
2. Merge Sorted Lists: We then merge list1 and list2 into a final result list. This is done using a standard two-pointer technique. We initialize a pointer for each list at the beginning. In a loop, we compare the elements at the current pointers, add the smaller element to result, and advance the pointer of the list from which the element was taken. This continues until one list is fully traversed. Finally, we append any remaining elements from the other list to result.

1class Solution {2    public List<Integer> getAllElements(TreeNode root1, TreeNode root2) {3        List<Integer> list1 = new ArrayList<>();4        inorder(root1, list1);5        6        List<Integer> list2 = new ArrayList<>();7        inorder(root2, list2);8        9        return merge(list1, list2);10    }11    12    private void inorder(TreeNode node, List<Integer> list) {13        if (node == null) {14            return;15        }16        inorder(node.left, list);17        list.add(node.val);18        inorder(node.right, list);19    }20    21    private List<Integer> merge(List<Integer> list1, List<Integer> list2) {22        List<Integer> result = new ArrayList<>();23        int p1 = 0, p2 = 0;24        25        while (p1 < list1.size() && p2 < list2.size()) {26            if (list1.get(p1) < list2.get(p2)) {27                result.add(list1.get(p1++));28            } else {29                result.add(list2.get(p2++));30            }31        }32        33        while (p1 < list1.size()) {34            result.add(list1.get(p1++));35        }36        37        while (p2 < list2.size()) {38            result.add(list2.get(p2++));39        }40        41        return result;42    }43}

• Initialize an empty result list result and two empty stacks, stack1 and stack2.
• Start a loop that continues as long as there are nodes to process in either tree (either stack is not empty or a current node pointer is not null).
• Inside the loop, for each tree, traverse as far left as possible from the current node, pushing each node onto its corresponding stack.
• After attempting to go left, compare the values of the nodes at the top of stack1 and stack2.
• If stack2 is empty or stack1's top element is smaller or equal, pop from stack1, add its value to result, and set the current node pointer for the first tree to the popped node's right child.
• Otherwise, do the same for stack2.
• Repeat until all nodes from both trees have been processed.
• Return result.

This approach simulates the in-order traversal iteratively for both trees at the same time and merges their outputs in a single pass.

1. Initialization: We use two stacks, s1 and s2, to manage the traversal for root1 and root2, respectively. We also have an empty result list.
2. Simultaneous Traversal and Merge: The main loop runs as long as either stack is non-empty or we still have nodes to visit in either tree. In each iteration: a. We push the path to the smallest unvisited node of the current subtree onto the stack. For root1, we keep pushing root1 to s1 and moving root1 = root1.left until root1 is null. We do the same for root2 and s2. b. Now, the top of s1 and s2 hold the next nodes in the in-order sequence for their respective trees. We compare their values. c. We pick the node with the smaller value. Let's say s1.peek().val is smaller. We pop this node from s1, add its value to result, and then set our current node pointer for the first tree to the popped node's right child. This is crucial because the next elements in the sequence are in the right subtree. d. If s2.peek().val was smaller (or if s1 was empty), we would do the same for s2.
3. This process continues until both trees are fully traversed, and the result list will contain all elements in sorted order.

1class Solution {2    public List<Integer> getAllElements(TreeNode root1, TreeNode root2) {3        List<Integer> result = new ArrayList<>();4        Stack<TreeNode> stack1 = new Stack<>();5        Stack<TreeNode> stack2 = new Stack<>();6 7        while (root1 != null || root2 != null || !stack1.isEmpty() || !stack2.isEmpty()) {8            // Go to the leftmost node in the current subtree for both trees9            while (root1 != null) {10                stack1.push(root1);11                root1 = root1.left;12            }13            while (root2 != null) {14                stack2.push(root2);15                root2 = root2.left;16            }17 18            // Compare the top elements of the stacks19            // If stack2 is empty or stack1's top is smaller/equal, process stack120            if (stack2.isEmpty() || (!stack1.isEmpty() && stack1.peek().val <= stack2.peek().val)) {21                root1 = stack1.pop();22                result.add(root1.val);23                root1 = root1.right; // Move to the right subtree to find the next element24            } else { // Otherwise, process stack225                root2 = stack2.pop();26                result.add(root2.val);27                root2 = root2.right; // Move to the right subtree28            }29        }30        return result;31    }32}