LeetCode（剑指 Offer）- 26. 树的子结构

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */

// 解决方案(1)
class Solution {

    private boolean ok = false;

    public boolean isSubStructure(TreeNode A, TreeNode B) {
        if (null == B) {
            return false;
        }

        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(A);
        TreeNode b = B;
        while (!queue.isEmpty()) {
            TreeNode a = queue.poll();
            if (a.val == b.val) {
                bfs(a, b);
                if (ok) {
                    break;
                } else {
                    b = B;
                }
            }
            if (a.left != null) {
                queue.offer(a.left);
            }
            if (a.right != null) {
                queue.offer(a.right);
            }
        }

        return ok;
    }

    private void bfs(TreeNode sa, TreeNode sb) {
        Queue<TreeNode> qa = new LinkedList<>();
        Queue<TreeNode> qb = new LinkedList<>();
        qa.offer(sa);
        qb.offer(sb);
        while (!qb.isEmpty()) {
            TreeNode a = qa.poll();
            TreeNode b = qb.poll();
            if (a == null || a.val != b.val) {
                return;
            }
            if (a.left != null) {
                qa.offer(a.left);
            }
            if (a.right != null) {
                qa.offer(a.right);
            }
            if (b.left != null) {
                qb.offer(b.left);
            }
            if (b.right != null) {
                qb.offer(b.right);
            }
        }
        ok = true;
    }
}

// 解决方案(2)
class Solution {
    public boolean isSubStructure(TreeNode A, TreeNode B) {
        return (A != null && B != null) && (recur(A, B) || isSubStructure(A.left, B) || isSubStructure(A.right, B));
    }
    boolean recur(TreeNode A, TreeNode B) {
        if(B == null) return true;
        if(A == null || A.val != B.val) return false;
        return recur(A.left, B.left) && recur(A.right, B.right);
    }
}

1
2
3
4
5
6
7
8
9
10
11
12
class Solution {
public:
    bool isSubStructure(TreeNode* A, TreeNode* B) {
        return (A != nullptr && B != nullptr) && (recur(A, B) || isSubStructure(A->left, B) || isSubStructure(A->right, B));
    }
private:
    bool recur(TreeNode* A, TreeNode* B) {
        if(B == nullptr) return true;
        if(A == nullptr || A->val != B->val) return false;
        return recur(A->left, B->left) && recur(A->right, B->right);
    }
};