Symmetric Tree

• LeetCode: Symmetric Tree

• Problem:

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

• Note:

Bonus points if you could solve it both recursively and iteratively.

• Examples:

For example, this binary tree [1,2,2,3,4,4,3] is symmetric:

    1
   / \
  2   2
 / \ / \
3  4 4  3

But the following [1,2,2,null,3,null,3] is not:

    1
   / \
  2   2
   \   \
   3    3

• Analysis:

The symmetric recursive rule from problem description is symmetric(left.left, right.right) && symmetric(left.right, right.left)

• Code - Java:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isSymmetric(TreeNode root) {
        if (root == null) {
            return true;
        }
        return helper(root.left, root.right);
    }
    private boolean helper(TreeNode left, TreeNode right) {
        if (left == null && right == null) {
            return true;
        }
        if (left == null || right == null) {
            return false;
        }
        if (left.val != right.val) {
            return false;
        }
        return helper(left.left, right.right) && helper(left.right, right.left);
    }
}

• Code - C++:

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool isSymmetric(TreeNode* root) {
    if (root == NULL) {
            return true;
        }
        return helper(root->left, root->right);
    }
    bool helper(TreeNode* left, TreeNode* right) {
        if (left == NULL && right == NULL) {
            return true;
        }
        if (left == NULL || right == NULL) {
            return false;
        }
        if (left->val != right->val) {
            return false;
        }
        return helper(left->left, right->right) && helper(left->right, right->left);
    }
};

• Code - Python:

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def isSymmetric(self, root):
        """
        :type root: TreeNode
        :rtype: bool
        """
        if root == None:
            return True
        return self.helper(root.left, root.right)
    def helper(self, left, right):
        if left == None and right == None:
            return True
        if left == None or right == None:
            return False
        if left.val != right.val:
            return False
        return self.helper(left.left, right.right) and self.helper(left.right, right.left)

• Code - C:

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
bool helper(struct TreeNode* left, struct TreeNode* right) {
    if (left == 0 && right == 0) {
        return true;
    }
    if (left == 0 || right == 0) {
        return false;
    }
    if (left->val != right->val) {
        return false;
    }
    return helper(left->left, right->right) && helper(left->right, right->left);
}
bool isSymmetric(struct TreeNode* root) {
    if (root == 0) {
        return true;
    }
    return helper(root->left, root->right);
}

• Time Complexity: O(logN), N is the number of nodes