Symmetric Tree | Leetcode

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
For example, this binary tree is symmetric:

    1
   / \
  2   2
 / \ / \
3  4 4  3

But the following is not:

    1
   / \
  2   2
   \   \
   3    3

Note:
Bonus points if you could solve it both recursively and iteratively.

Confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

---

Solution: Recurse check for equality, except that we compare the left.left child with the right.right child, and the right.left child with the left.right child.

Algorithm 1: Recursive (easier)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean isSymmetric(TreeNode root) {
        if (root == null) return true;
        return isSymmetric(root.left, root.right);
    }
    
    private boolean isSymmetric(TreeNode left, TreeNode right) {
        if (left == right) return true;
        if (left == null || right == null) return false;
        if (left.val != right.val) return false;
        return isSymmetric(left.left, right.right) && isSymmetric(right.left, left.right);
    }
}

Algorithm 2: Iterative (harder)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    private static class TreeNodePair {
        private TreeNode node1;
        private TreeNode node2;
        
        private TreeNodePair(TreeNode node1, TreeNode node2) {
            this.node1 = node1;
            this.node2 = node2;
        }
    }

    public boolean isSymmetric(TreeNode root) {
        if (root == null) return true;
        Stack stack = new Stack<>();
        stack.add(new TreeNodePair(root.left, root.right));
        do {
            TreeNodePair pair = stack.pop();
            if (pair.node1 == pair.node2) continue;
            if (pair.node1 == null || pair.node2 == null) return false;
            if (pair.node1.val != pair.node2.val) return false;
            stack.add(new TreeNodePair(pair.node1.left, pair.node2.right));
            stack.add(new TreeNodePair(pair.node2.left, pair.node1.right));
        } while (!stack.isEmpty());
        return true;
    }
}