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.

Thoughts:

1. Recursively: each call needs to check the current two nodes value and need to recursively call two pairs: (left.left, right.right) and (left.right, right.left).

2. Iteratively: Same logic; Using stack, each time check the and push the corresponding pairs of child node into the stack

Code: Recursive

public boolean isSymmetric(TreeNode root) {
    return root==null || isSymmetricHelp(root.left, root.right);
}

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

Code: Iterative

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isSymmetric(TreeNode root) {
        if(root == null) return true;
        Stack<TreeNode> s = new Stack<TreeNode>();
        s.push(root.left);
        s.push(root.right); // relative ordering in the pair does not matter
        TreeNode l, r;

        while(!s.empty()){
            l = s.pop(); r = s.pop();
            if(l == null && r == null) continue;
            if(l == null || r == null || l.val != r.val) return false;
            s.push(l.left); s.push(r.right); // relative ordering in the pair does not matter
            s.push(l.right); s.push(r.left);
        }


        return true;
    }
}