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This approach uses recursion to compare the left subtree and right subtree of the binary tree. For two trees to be mirror images, the following three conditions must be true:
Recursion is an elegant way to check this condition for each node in the tree.
Time Complexity: O(n), where n is the number of nodes in the binary tree, since we traverse every node once.
Space Complexity: O(h), where h is the height of the tree, due to the recursion call stack.
1public class TreeNode {
2 public int val;
3 public TreeNode left;
4 public TreeNode right;
5 public TreeNode(int x) { val = x; }
6}
7
8public class Solution {
9 public bool IsSymmetric(TreeNode root) {
10 return IsMirror(root, root);
11 }
12
13 private bool IsMirror(TreeNode t1, TreeNode t2) {
14 if (t1 == null && t2 == null) return true;
15 if (t1 == null || t2 == null) return false;
16 return (t1.val == t2.val) &&
17 IsMirror(t1.right, t2.left) &&
18 IsMirror(t1.left, t2.right);
19 }
20}This C# solution defines a Solution class with a public method IsSymmetric. The helper method IsMirror checks if two subtrees are mirrors using recursion.
This approach utilizes a queue data structure to iteratively compare nodes in the binary tree. It behaves like the recursive method mirroring trees, but it exchanges recursion for a loop that dequeues two nodes at a time.
Time Complexity: O(n), n being indicative of the number of nodes enumerated.
Space Complexity: O(n), underscored by the queue's need for storing nodes in tandem with iteration.
This Java method formulates symmetry checking via iterative processing using the Queue interface. Two nodes are dequeued and compared in each iteration featuring equality checks for left and right subtree counterparts.