首页 > 代码库 > 101. Symmetric Tree 二叉树是否对称

101. Symmetric Tree 二叉树是否对称

2024-10-08 22:36:39   218人阅读

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.

  1. /**
  2.  * Definition for a binary tree node.
  3.  * public class TreeNode {
  4.  *     public int val;
  5.  *     public TreeNode left;
  6.  *     public TreeNode right;
  7.  *     public TreeNode(int x) { val = x; }
  8.  * }
  9.  */
  10. public class Solution {
  11.     public bool IsSymmetric(TreeNode root) {
  12. 		if (root == null) return true;
  13. 		Queue<TreeNode> queue = new Queue<TreeNode>();
  14. 		queue.Enqueue(root);
  15. 		while (queue.Count > 0) {
  16. 			int count = queue.Count;
  17. 			List<int?> nextLevel = new List<int?>();
  18. 			for (int i = 0; i < count; i++) {
  19. 				TreeNode curNode = queue.Dequeue();
  20. 				if (curNode.left != null) {
  21. 					queue.Enqueue(curNode.left);
  22. 					nextLevel.Add(curNode.left.val);
  23. 				} else {
  24. 					nextLevel.Add(null);
  25. 				}
  26. 				if (curNode.right != null) {
  27. 					queue.Enqueue(curNode.right);
  28. 					nextLevel.Add(curNode.right.val);
  29. 				} else {
  30. 					nextLevel.Add(null);
  31. 				}
  32. 			}
  33. 			if (!LevelIsSymmetric(nextLevel)){
  34. 				return false;
  35. 			}
  36. 		}
  37. 		return true;
  38.     }
  39.     
  40. 	public bool LevelIsSymmetric(List<int?> list) {
  41. 		int left = 0;
  42. 		int right = list.Count - 1;
  43. 		while (right > left) {
  44. 			if (list[left++] != list[right--]) {
  45. 				return false;
  46. 			}
  47. 		}
  48. 		return true;
  49. 	}
  50. }






null


101. Symmetric Tree 二叉树是否对称


联系
我们