Construct Binary Tree from Preorder and Inorder
Given preorder and inorder traversal arrays, construct the binary tree.
Output the level-order traversal of the constructed tree, space-separated.
Example:
3,9,20,15,7 9,3,15,20,7
3 9 20 null null 15 7
- The preorder traversal array
[3, 9, 20, 15, 7]represents the order in which nodes are visited: root, left subtree, right subtree. The first element3is the root node. - The inorder traversal array
[9, 3, 15, 20, 7]represents the order in which nodes are visited: left subtree, root, right subtree. This helps to identify the left and right child nodes of the root3, which are9and20respectively. - The left subtree of
3only contains9, and the right subtree of3is constructed from the remaining elements[15, 20, 7]in the inorder array, with20as the root,15as its left child, and7as its right child. - The constructed binary tree is then traversed level-by-level to produce the output:
3(root),9(left child of3),20(right child of3),null(no left child of9),null(no right child of9),15(left child of20),7(right child of20).
Constraints:
- 1 <= length <= 3000
- All values unique
- -3000 <= values <= 3000
Background Knowledge
The problem involves constructing a binary tree from given preorder and inorder traversal arrays. To tackle this, it's essential to understand the concepts of tree traversals. In a preorder traversal, we visit the root node first, then recursively traverse the left subtree, and finally the right subtree. On the other hand, in an inorder traversal, we visit the left subtree, then the root node, and finally the right subtree. These traversal techniques are fundamental in tree data structures.
Understanding the properties of binary trees is also crucial. A binary tree is a tree data structure in which each node has at most two children, referred to as the left child and the right child. The root node is the topmost node in the tree. The height of a tree is the number of edges on the longest path from the root to a leaf node. Familiarity with tree-related terminology, such as node, edge, leaf node, and internal node, will help in visualizing and solving the problem.
To construct a binary tree from the given traversals, we need to identify the root node and then recursively construct the left and right subtrees. This process involves finding the correct positions of nodes in the inorder traversal based on the preorder traversal. The relationship between the preorder and inorder traversals will be key in solving this problem. We will use this relationship to identify the root node and the boundaries of the left and right subtrees in the inorder traversal.
Algorithm/Approach
The general approach to solve this type of problem involves using a recursive or iterative method to construct the binary tree. We will utilize the properties of preorder and inorder traversals to identify the root node and the boundaries of the left and right subtrees. This approach will involve:
- Identifying the root node from the preorder traversal
- Finding the left and right subtrees in the inorder traversal
- Recursively constructing the left and right subtrees
Step-by-Step Strategy
To implement the solution, follow these steps:
- Identify the root node from the preorder traversal.
- Find the index of the root node in the inorder traversal.
- Determine the boundaries of the left and right subtrees in the inorder traversal.
- Recursively construct the left and right subtrees using the same approach.
- Combine the constructed left, root, and right subtrees to form the final binary tree.
- Perform a level-order traversal on the constructed tree to obtain the output.
Common Pitfalls
When implementing the solution, watch out for:
- Incorrectly identifying the root node or the boundaries of the left and right subtrees.
- Failing to handle edge cases, such as an empty preorder or inorder traversal.
- Not properly recursively constructing the left and right subtrees.
Time & Space Complexity
The expected time complexity for this problem is O(n), where n is the number of nodes in the tree, since we need to process each node once. The space complexity is also O(n), as in the worst case, the recursive call stack can go up to the height of the tree, which is n for an unbalanced tree. However, for a balanced tree, the space complexity would be O(logn) due to the recursive call stack.