Invert Binary Tree
Given the root of a binary tree (as a level-order array), invert it (mirror/flip left and right children recursively).
Output the level-order traversal of the inverted tree as space-separated values.
Example:
4,2,7,1,3,6,9
4 7 2 9 6 3 1
- The given binary tree has a root node with value 4, and its left and right children are 2 and 7, respectively.
- To invert the tree, we swap the left and right children of each node recursively: the left child of the root becomes 7 (with children 6 and 9), and the right child becomes 2 (with children 3 and 1).
- We then perform a level-order traversal of the inverted tree, visiting nodes in the order: root (4), children (7, 2), grandchildren (9, 6, 3, 1).
- The final output is the level-order traversal of the inverted tree as space-separated values: 4 7 2 9 6 3 1
Constraints:
- 0 <= number of nodes <= 100
- -100 <= Node.val <= 100
Background Knowledge
The problem involves working with a binary tree, which is a data structure in which each node has at most two children (i.e., left child and right child). In this case, we're given the root of the binary tree as a level-order array, which means the nodes are arranged in a specific order: the root node, followed by its children, then its grandchildren, and so on. To invert a binary tree means to mirror or flip its left and right children recursively. This operation requires a good understanding of recursion, which is a programming technique where a function calls itself repeatedly until it reaches a base case.
Understanding the structure of a binary tree is crucial to solving this problem. Each node in the tree has a value and references to its left and right children. When inverting the tree, we need to swap the left and right children of each node. This process needs to be repeated recursively for all nodes in the tree. The level-order traversal of a binary tree is also an important concept, as it involves visiting all nodes at a given level before moving on to the next level. In this case, we're asked to output the level-order traversal of the inverted tree as space-separated values.
The problem requires a combination of tree traversal and recursion to solve. Tree traversal involves visiting each node in the tree in a specific order, while recursion allows us to repeat this process for all nodes in the tree. By understanding how to traverse a binary tree and how to use recursion to repeat this process, we can develop an effective solution to invert the tree.
Algorithm/Approach
The general approach to solving this problem involves using a recursive algorithm to traverse the binary tree and invert its nodes. We can use a depth-first search (DFS) or breadth-first search (BFS) approach to traverse the tree, but in this case, a recursive DFS approach is more suitable. The algorithm will involve swapping the left and right children of each node and repeating this process recursively for all nodes in the tree.
Step-by-Step Strategy
To implement the solution, follow these steps:
- Create a function to invert the binary tree recursively
- Within the function, swap the left and right children of the current node
- Recursively call the function on the left and right children of the current node
- Use a level-order traversal to output the values of the inverted tree as space-separated values
- Start the traversal from the root node and visit all nodes at each level before moving on to the next level
Common Pitfalls
When implementing the solution, watch out for the following common pitfalls:
- Forgetting to handle the base case for recursion (i.e., when the current node is null)
- Not properly swapping the left and right children of each node
- Not using recursion correctly to repeat the process for all nodes in the tree
- Not implementing the level-order traversal correctly to output the values of the inverted tree
Time & Space Complexity
The expected time complexity for this problem is O(n), where n is the number of nodes in the binary tree, since we need to visit each node once to invert the tree. The expected space complexity is O(n) as well, since in the worst case, the recursive call stack can go as high as the height of the tree, which is n for an unbalanced tree. However, for a balanced tree, the space complexity would be O(logn), where logn is the height of the tree.