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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.
1function TreeNode(val) {
2 this.val = val;
3 this.left = this.right = null;
4}
5
6var isSymmetric = function(root) {
7 return isMirror(root, root);
8};
9
10function isMirror(t1, t2) {
11 if (!t1 && !t2) return true;
12 if (!t1 || !t2) return false;
13 return (t1.val === t2.val) &&
14 isMirror(t1.right, t2.left) &&
15 isMirror(t1.left, t2.right);
16}In JavaScript, the function isSymmetric initiates the recursion by calling isMirror with the root node provided as both arguments. This function verifies if the tree nodes supplied as arguments mirror each other.
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.