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Problem Statement
Write Recursive Approach for Inorder Traversal of Binary Tree.
Given a binary tree, write a recursive algorithm to perform an inorder traversal of the tree and return the elements in the order they were visited.
Examples
Example 1:
Input: 5 / \ 3 8 / \ / \ 2 4 7 9 Output: Inorder Traversal: [2, 3, 4, 5, 7, 8, 9] Explanation: The binary tree has the following structure: 5 / \ 3 8 / \ / \ 2 4 7 9 Performing an inorder traversal visits the nodes in ascending order: 2, 3, 4, 5, 7, 8, 9.
Example 2:
Input: 10 / \ 6 15 / \ / \ 3 8 12 18 \ 9 Output: Inorder Traversal: [3, 6, 8, 9, 10, 12, 15, 18] Explanation: The binary tree has the following structure: 10 / \ 6 15 / \ / \ 3 8 12 18 \ 9 Performing an inorder traversal visits the nodes in ascending order: 3, 6, 8, 9, 10, 12, 15, 18.
Example 3:
Input: 20 / \ 12 25 / \ / \ 8 15 22 28 / \ \ 6 10 24 Output: Inorder Traversal: [6, 8, 10, 12, 15, 20, 22, 24, 25, 28] Explanation: The binary tree has the following structure: 20 / \ 12 25 / \ / \ 8 15 22 28 / \ \ 6 10 24 Performing an inorder traversal visits the nodes in ascending order: 6, 8, 10, 12, 15, 20, 22, 24, 25, 28.
Constraints:
- The number of nodes in the tree is in the range
[0, 100]
. -100 <= Node.val <= 100
Solution
The inorder traversal of a binary tree follows a recursive approach. We can define the algorithm as follows:
- Base Case: If the current node is null, return.
- Recursive Case:
- Recursively traverse the left subtree.
- Visit the current node.
- Recursively traverse the right subtree.
By following this approach, we can perform an inorder traversal of the binary tree.
Code
Here is the code for this algorithm:
Time and Space Complexity
The time complexity of the algorithm is O(N), where n is the number of nodes in the binary tree. This is because we need to visit each node exactly once.
The space complexity of the algorithm is O(h), where h is the height of the binary tree. In the worst case, where the binary tree is skewed (i.e., a linked list), the height h is equal to the number of nodes, resulting in O(N) space complexity. However, in a balanced binary tree, the height h is logarithmic to the number of nodes, resulting in efficient space usage with O(log N) space complexity.
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