Symmetric Tree 🌟

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

O(N) Time solution

• we will traverse left and right subtrees of the root with the same type of traversal.
• we compare the value of left with right or value of right with left , if they are not equal we return false.
• we recurse for left’s left with right’s right and left’s right with right’s left.

Code

class Solution {
public:
    bool isSymmetric(TreeNode* root){
        return root == NULL || isSymmetricHelp(root->left, root->right);
    }

    bool 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);
    }
};

O(N) Time, using 2 queue, iterative solution

• Same recursive solution can be converted to iterative solution by using queue.
• Remember while using 2 queue we push left->left,left->right in 1st queue and right->right,right->left in 2nd queue.

class Solution {
public:
    bool isSymmetric(TreeNode* root){
        if (!root) return true;

        TreeNode *left, *right;
        queue<TreeNode*> q1, q2;
        q1.push(root->left);
        q2.push(root->right);

        while (!q1.empty() && !q2.empty()) {
            left = q1.front();
            q1.pop();
            right = q2.front();
            q2.pop();

            if (left == NULL && right == NULL) continue;
            if (left == NULL || right == NULL) return false;
            if (left->val != right->val) return false;

            q1.push(left->left);
            q1.push(left->right);
            q2.push(right->right);
            q2.push(right->left);
        }
        return true;
    }
};

O(N) Time, using 1 queue, iterative solution

• We can use 1 queue instead of 2.
• remember that while using 1 queue we do left->left,right->right,left->right,right->left.

class Solution{
public:
    bool isSymmetric(TreeNode *root){
        TreeNode *left, *right;
        if (!root) return true;

        queue<TreeNode *> q;
        q.push(root->left);
        q.push(root->right);

        while (!q.empty()){
            left = q.front();
            q.pop();
            right = q.front();
            q.pop();

            if (left == NULL && right == NULL) continue;
            if (left == NULL || right == NULL) return false;
            if (left->val != right->val) return false;

            q.push(left->left);
            q.push(right->right);
            q.push(left->right);
            q.push(right->left);
        }
        return true;
    }
};