101. Symmetric Tree LeetCode Solution

In this guide, you will get 101. Symmetric Tree LeetCode Solution with the best time and space complexity. The solution to Symmetric Tree problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Table of Contents

  1. Problem Statement
  2. Complexity Analysis
  3. Symmetric Tree solution in C++
  4. Symmetric Tree solution in Java
  5. Symmetric Tree solution in Python
  6. Additional Resources
101. Symmetric Tree LeetCode Solution image

Problem Statement of 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

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?

Complexity Analysis

  • Time Complexity: O(n)
  • Space Complexity: O(h)

101. Symmetric Tree LeetCode Solution in C++

class Solution {
 public:
  bool isSymmetric(TreeNode* root) {
    return isSymmetric(root, root);
  }

 private:
  bool isSymmetric(TreeNode* p, TreeNode* q) {
    if (!p || !q)
      return p == q;

    return p->val == q->val &&                //
           isSymmetric(p->left, q->right) &&  //
           isSymmetric(p->right, q->left);
  }
};
/* code provided by PROGIEZ */

101. Symmetric Tree LeetCode Solution in Java

class Solution {
  public boolean isSymmetric(TreeNode root) {
    return isSymmetric(root, root);
  }

  private boolean isSymmetric(TreeNode p, TreeNode q) {
    if (p == null || q == null)
      return p == q;

    return p.val == q.val && isSymmetric(p.left, q.right) && isSymmetric(p.right, q.left);
  }
}
// code provided by PROGIEZ

101. Symmetric Tree LeetCode Solution in Python

class Solution:
  def isSymmetric(self, root: TreeNode | None) -> bool:
    def isSymmetric(p: TreeNode | None, q: TreeNode | None) -> bool:
      if not p or not q:
        return p == q
      return (p.val == q.val and
              isSymmetric(p.left, q.right) and
              isSymmetric(p.right, q.left))

    return isSymmetric(root, root)
# code by PROGIEZ

Additional Resources

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