LeetCode: isSameTree1 解题报告

isSameTree1

Given two binary trees, write a function to check if they are equal or not.

Two binary trees are considered equal if they are structurally identical and the nodes have the same value.

SOLUTION 1 & SOLUTION 2:

以下是递归及非递归解法:

1. 递归解法就是判断当前节点是不是相同，及左右子树是不是相同树

2. 非递归解法使用了先根遍历。这个算法比较简单一点

 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 isSameTree1(TreeNode p, TreeNode q) {

13         if (p == null && q == null) {

14             return true;

15         }

16         

17         if (p == null || q == null) {

18             return false;

19         }

20         

21         return p.val == q.val &&

22            isSameTree(p.left, q.left) &&

23            isSameTree(p.right, q.right);

24     }

25     

26     // Solution 2:

27     public boolean isSameTree(TreeNode p, TreeNode q) {

28         if (p == null && q == null) {

29             return true;

30         }

31         

32         if (p == null || q == null) {

33             return false;

34         }

35         

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

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

38         

39         s1.push(p);

40         s2.push(q);

41         

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

43             TreeNode cur1 = s1.pop();

44             TreeNode cur2 = s2.pop();

45             

46             // 弹出的节点的值必须相等 

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

48                 return false;

49             }

50             

51             // tree1的right节点，tree2的right节点，可以同时不为空，也可以同时为空，否则返回false.

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

53                 s1.push(cur1.left);

54                 s2.push(cur2.left);

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

56                 return false;

57             }

58             

59             // tree1的左节点，tree2的left节点，可以同时不为空，也可以同时为空，否则返回false.

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

61                 s1.push(cur1.right);

62                 s2.push(cur2.right);

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

64                 return false;

65             }

66         }

67         

68         return true;

69     }

70 }

View Code

 

GITHUB:

https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/tree/IsSameTree1.java