Symmetric Tree LeetCode Solution | Easy Approach

Minimum Cost to Merge Stones

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


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

Symmetric Tree Solutions

Time: O(n)
Space: O(h)


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

  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);


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);


class Solution:
  def isSymmetric(self, root: Optional[TreeNode]) -> bool:
    def isSymmetric(p: Optional[TreeNode], q: Optional[TreeNode]) -> 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)

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