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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.
1class TreeNode {
2 int val;
3 TreeNode left;
4 TreeNode right;
5 TreeNode(int x) { val = x; }
6}
7
8public class Solution {
9 public boolean isSymmetric(TreeNode root) {
10 return isMirror(root, root);
11 }
12
13 private boolean 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}In the Java solution, class Solution uses a private helper method isMirror to check if two subtrees are mirror images. The public method isSymmetric initiates this check with the root of the binary tree.
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.
public class TreeNode {
public int val;
public TreeNode left;
public TreeNode right;
public TreeNode(int x) { val = x; }
}
public class Solution {
public bool IsSymmetric(TreeNode root) {
if (root == null) return true;
Queue<TreeNode> queue = new Queue<TreeNode>();
queue.Enqueue(root);
queue.Enqueue(root);
while (queue.Count != 0) {
TreeNode t1 = queue.Dequeue();
TreeNode t2 = queue.Dequeue();
if (t1 == null && t2 == null) continue;
if (t1 == null || t2 == null) return false;
if (t1.val != t2.val) return false;
queue.Enqueue(t1.left);
queue.Enqueue(t2.right);
queue.Enqueue(t1.right);
queue.Enqueue(t2.left);
}
return true;
}
}Utilizing Queue from the C# library, this solution iteratively evaluates a tree's symmetry. Nodes are enqueued and dequeued in pairs to assess comparability, verifying left and right elements throughout.