Symmetric Tree LeetCode Solution Leetcode Solution

Difficulty Level Easy
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Problem Statement

The Symmetric Tree LeetCode Solution – “Symmetric Tree” states that given the root of the binary tree and we need to check if the given binary tree is a mirror of itself(symmetric around its center) or not? If Yes, we need to return true otherwise, false.

Example:

Symmetric Tree LeetCode Solution Leetcode Solution

 

Note that nodes with the same color have the same node value. The binary tree is a Symmetric Tree.

 

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

Explanation:

  • Check the above diagram for a better understanding.
Input:  root = [1,2,2,null,3,null,3]
Output: false

Symmetric Tree LeetCode Solution Leetcode Solution

Explanation:

  • Check the above diagram for a better understanding.

Approach

Idea:

  1. The main idea to solve this problem is to use Recursion.
  2. Now, a tree is called symmetric if the left subtree must be a mirror reflection of the right subtree.
  3. Also, Two trees are said to be a mirror with respect to each other if:
    1. Their roots are of the same value.
    2. The right subtree of each tree is a mirror reflection of the left subtree of another tree.
  4. So, perform the recursion with the following cases:
    1. For the base case:
      • If both root nodes are null pointers, return true.
      • If exactly one of them is a null node, return false.
    2. In general:
      • Root nodes must have the same value and,
      • The left subtree and right subtree must be the mirror reflection with respect to each other.
      • So, return true if the root node’s values are the same and left as well as right subtrees are symmetric.

Code

Symmetric Tree Leetcode C++ Solution:

class Solution {
public:
    bool check(TreeNode* root1,TreeNode* root2){
        if(root1==nullptr or root2==nullptr){
            return root1==root2;
        }
        return root1->val==root2->val and check(root1->left,root2->right) and check(root1->right,root2->left);
    }
    bool isSymmetric(TreeNode* root) {
        return check(root,root);
    }
};

Symmetric Tree Leetcode Java Solution:

class Solution {
    private boolean check(TreeNode root1,TreeNode root2){
        if(root1==null || root2==null){
            return root1==root2;
        }
        return root1.val==root2.val && check(root1.left,root2.right) && check(root1.right,root2.left);
    }
    public boolean isSymmetric(TreeNode root) {
        return check(root,root);
    }
}

Complexity Analysis for Symmetric Tree Leetcode Solution

Time Complexity

The time complexity of the above code is O(N) since we traverse the entire input tree once where N = the total number of nodes in the tree.

Space Complexity

The space complexity of the above code is O(N) [considering recursive calls also]. The number of recursive calls is bounded by the height of the tree and in the worst case, the tree can be linear. Hence, Space Complexity is O(N).

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