Leetcode 0101. Symmetric Tree

Given the `root` of a binary tree, check whether it is a mirror of itself (i.e., symmetric around its center).　 Example 1:　 Input: root = [1,2,2,3,4,4,3], Output: true　 Example 2:　 Input: root = [1,2,2,null,3,null,3], Output: false　

Easy

Description

Given the root of a binary tree, check whether it is a mirror of itself (i.e., symmetric around its center).

Example 1:

Input: root = [1,2,2,3,4,4,3]
Output: true

Example 2:

Input: root = [1,2,2,null,3,null,3]
Output: false

Constraints:

• The number of nodes in the tree is in the range [1, 1000].
• -100 <= Node.val <= 100

Follow up

• Could you solve it both recursively and iteratively?

Solution

/**
 * 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 {

  List<Integer> left = new ArrayList<>();
  List<Integer> right = new ArrayList<>();

  /**
   * Depth First Traversal: preorder
   * Node -> Left -> Right
   */
  public void preorder(TreeNode root, List<Integer> list) {
    if (root == null) {
      list.add(null);
      return;
    }
    list.add(root.val);
    preorder(root.left, list);
    preorder(root.right, list);
  }

  public void mirroredPeorder(TreeNode root, List<Integer> list) {
    if (root == null) {
      list.add(null);
      return;
    }
    list.add(root.val);
    mirroredPeorder(root.right, list);
    mirroredPeorder(root.left, list);
  }

  /**
   * Recursion
   * Time Complexity: BigO(n)
   * Space Complexity: BigO(n)
   */
  public boolean isSymmetric(TreeNode root) {
    preorder(root.left, left);
    mirroredPeorder(root.right, right);
    return left.equals(right);
  }
}