LeetCode: Symmetric Tree 解题报告

Symmetric Tree

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

For example, this binary tree is symmetric:

    1
   / \
  2   2
 / \ / \
3  4 4  3
But the following is not:
    1
   / \
  2   2
   \   \
   3    3
Note:
Bonus points if you could solve it both recursively and iteratively.

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

SOL 1 & 2:

两个解法，递归和非递归.

 1 /**

 2  * Definition for binary tree

 3  * public class TreeNode {

 4  *     int val;

 5  *     TreeNode left;

 6  *     TreeNode right;

 7  *     TreeNode(int x) { val = x; }

 8  * }

 9  */

10 public class Solution {

11     // solution 1:

12     public boolean isSymmetric(TreeNode root) {

13         if (root == null) {

14             return true;

15         }

16         

17         return isSymmetricTree(root.left, root.right);

18     }

19     

20     /*

21      * 判断两个树是否互相镜像 

22         (1) 根必须同时为空，或是同时不为空

23      * 

24      * 如果根不为空: 

25         (1).根的值一样 

26         (2).r1的左树是r2的右树的镜像 

27         (3).r1的右树是r2的左树的镜像

28      */

29     public boolean isSymmetricTree1(TreeNode root1, TreeNode root2) {

30         if (root1 == null && root2 == null) {

31             return true;

32         }

33         

34         if (root1 == null || root2 == null) {

35             return false;

36         }

37         

38         return root1.val == root2.val && 

39              isSymmetricTree(root1.left, root2.right) && 

40              isSymmetricTree(root1.right, root2.left);

41     }

42     

43     // solution 2:

44     public boolean isSymmetricTree(TreeNode root1, TreeNode root2) {

45         if (root1 == null && root2 == null) {

46             return true;

47         }

48         

49         if (root1 == null || root2 == null) {

50             return false;

51         }

52         

53         Stack<TreeNode> s1 = new Stack<TreeNode>();

54         Stack<TreeNode> s2 = new Stack<TreeNode>();

55         

56         // 一定记得初始化

57         s1.push(root1);

58         s2.push(root2);

59         

60         while (!s1.isEmpty() && !s2.isEmpty()) {

61             TreeNode cur1 = s1.pop();

62             TreeNode cur2 = s2.pop();

63             

64             if (cur1.val != cur2.val) {

65                 return false;

66             }

67             

68             if (cur1.left != null && cur2.right != null) {

69                 s1.push(cur1.left);

70                 s2.push(cur2.right);

71             } else if (!(cur1.left == null && cur2.right == null)) {

72                 return false;

73             }

74             

75             if (cur1.right != null && cur2.left != null) {

76                 s1.push(cur1.right);

77                 s2.push(cur2.left);

78             } else if (!(cur1.right == null && cur2.left == null)) {

79                 return false;

80             }

81         }

82         

83         return true;

84     }

85 }

View Code

 

2015.1.20:

简化了递归的解法，root与root本身是一对对称树。

 1 /**

 2  * Definition for binary tree

 3  * public class TreeNode {

 4  *     int val;

 5  *     TreeNode left;

 6  *     TreeNode right;

 7  *     TreeNode(int x) { val = x; }

 8  * }

 9  */

10 public class Solution {

11     public boolean isSymmetric(TreeNode root) {

12         return isMirror(root, root);

13     }

14     

15     public boolean isMirror(TreeNode r1, TreeNode r2) {

16         if (r1 == null && r2 == null) {

17             return true;

18         } else if (r1 == null || r2 == null) {

19             return false;

20         }

21         

22         return r1.val == r2.val

23              && isMirror(r1.left, r2.right)

24              && isMirror(r1.right, r2.left);

25     }

26 }

View Code

 

GITHUB:

https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/tree/IsSymmetric.java