101. Symmetric Tree

Problem

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.

Solution (Java)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public boolean isSymmetric(TreeNode root) {
        if (root == null) {
            return true;
        }
        return helper(root.left, root.right);
    }

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

Solution 1 (Javascript)

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @return {boolean}
 */
var isSymmetric = function(root) {
  if (!root) return true;
  return helper(root.left, root.right);
};

var helper = function (p, q) {
  if ((!p && q) || (p && !q) || (p && q && p.val !== q.val)) return false;
  if (p && q) return helper(p.left, q.right) && helper(p.right, q.left);
  return true;
};

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(1).

Solution 2

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @return {boolean}
 */
var isSymmetric = function(root) {
  if (!root) return true;

  var p = [root.left];
  var q = [root.right];
  var ll = null;
  var rr = null;

  while (p.length && q.length) {
    ll = p.pop();
    rr = q.pop();

    if (!ll && !rr) continue;
    if (!ll || !rr) return false;
    if (ll.val !== rr.val) return false;

    p.push(ll.left);
    p.push(ll.right);
    q.push(rr.right);
    q.push(rr.left);
  }

  return true;
};

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).