tsanmmmm2000's Tech Blog

Question

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

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

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

Thinking

此題可以用深度優先搜尋法(Depth-first Search)，採recursive的方式；或是廣度優先搜尋法(Breadth-first Search)，利用Queue以iterative的方式處理

My C# Solution

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

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public bool IsSymmetric(TreeNode root) {
        if (root == null) return true;
        
        var queue = new Queue<TreeNode>();
        queue.Enqueue(root.left);
        queue.Enqueue(root.right);
        while(queue.Count > 0)
        {
            var right = queue.Dequeue();
            var left = queue.Dequeue();
            if (right == null && left == null) continue;
            if (right == null 
               || left == null 
               || right.val != left.val) return false;

            queue.Enqueue(left.left);
            queue.Enqueue(right.right);
            queue.Enqueue(left.right);
            queue.Enqueue(right.left);
        }
        return true;
    }
}