101. Symmetric Tree

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

For example, this binary tree[1,2,2,3,4,4,3]is symmetric:

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

But the following[1,2,2,null,3,null,3]is not:

    1
   / \
  2   2
   \   \
   3    3

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

Analysis:

Recursive approach:

Create a boolean dualTraverse method to take in 2 nodes for traverse. They traverse in the exact opposite direction.

Complexity:

Time: O(N)

Space: O(logn)

Code:

Recursive Approach:

/**
 * 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 dualTraverse(root.left, root.right);
    }

    private boolean dualTraverse(TreeNode rl, TreeNode rr)
    {
        if(rl == null && rr == null)
        {
            return true;
        }
        if(rl == null || rr == null || rl.val != rr.val)
        {
            return false;
        }
        return dualTraverse(rl.left, rr.right) && dualTraverse(rl.right, rr.left);
    }
}

Iterative approach: