Given the root of a binary tree, invert the tree, and return its root.
Example 1:
Input: root = [4,2,7,1,3,6,9]
Output: [4,7,2,9,6,3,1]
Example 2:
Input: root = [2,1,3]
Output: [2,3,1]
Example 3:
Input: root = []
Output: []
Constraints:
- The number of nodes in the tree is in the range
[0, 100]. -100 <= Node.val <= 100
Approach - Recursion
Time: O(n) Space: O(n)
class Solution {
public TreeNode invertTree(TreeNode root) {
if (root == null)
return null;
TreeNode t = root.right;
root.right = root.left;
root.left = t;
invertTree(root.left);
invertTree(root.right);
return root;
}
}Approach - BFS
- We will BFS and use the simple reverse logic
- For loop exist to check for one level so it is required here but might not be required in other solutions
Time: O(n) Space: O(n)
/**
* 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 TreeNode invertTree(TreeNode root) {
if (root == null) return null;
Queue<TreeNode> q = new LinkedList<>();
q.add(root);
while(!q.isEmpty()) {
TreeNode n = q.poll();
TreeNode t = n.left;
n.left = n.right;
n.right = t;
if (n.left != null) q.add(n.left);
if (n.right != null) q.add(n.right);
}
return root;
}
}