本文实例讲述了C++实现寻找最低公共父节点的方法,是数据结构中二叉树的经典算法。分享给大家供大家参考。具体方法如下:
最低公共父节点,意思很好理解。
思路1:最低公共父节点满足这样的条件:两个节点分别位于其左子树和右子树,那么定义两个bool变量,leftFlag和rightFlag,如果在左子树中,leftFlag为true,如果在右子树中,rightFlag为true,仅当leftFlag == rightFlag == true时,才能满足条件。
实现代码如下:
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
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
|
#include <iostream> using namespace std; struct Node { Node( int i = 0, Node *pLeft = NULL, Node *pRight = NULL) : data(i), left(pLeft), right(pRight) {} Node *left; Node *right; int data; }; Node *constructNode(Node **pNode1, Node **pNode2) { Node *node12 = new Node(12); Node *node11 = new Node(11); Node *node10 = new Node(10); Node *node9 = new Node(9, NULL, node12); Node *node8 = new Node(8, node11, NULL); Node *node7 = new Node(7); Node *node6 = new Node(6); Node *node5 = new Node(5, node8, node9); Node *node4 = new Node(4, node10); Node *node3 = new Node(3, node6, node7); Node *node2 = new Node(2, node4, node5); Node *node1 = new Node(1, node2, node3); *pNode1 = node6; *pNode2 = node12; return node1; } bool isNodeIn(Node *root, Node *node1, Node *node2) { if (node1 == NULL || node2 == NULL) { throw ( "invalid node1 and node2" ); return false ; } if (root == NULL) return false ; if (root == node1 || root == node2) { return true ; } else { return isNodeIn(root->left, node1, node2) || isNodeIn(root->right, node1, node2); } } Node *lowestFarther(Node *root, Node *node1, Node *node2) { if (root == NULL || node1 == NULL || node2 == NULL || node1 == node2) { return NULL; } bool leftFlag = false ; bool rightFlag = false ; leftFlag = isNodeIn(root->left, node1, node2); rightFlag = isNodeIn(root->right, node1, node2); if (leftFlag == true && rightFlag == true ) { return root; } else if (leftFlag == true ) { return lowestFarther(root->left, node1, node2); } else { return lowestFarther(root->right, node1, node2); } } void main() { Node *node1 = NULL; Node *node2 = NULL; Node *root = constructNode(&node1, &node2); cout << "node1: " << node1->data << endl; cout << "node2: " << node2->data << endl; cout << "root: " << root->data << endl; Node *father = lowestFarther(root, node1, node2); if (father == NULL) { cout << "no common father" << endl; } else { cout << "father: " << father->data << endl; } } |
这类问题在面试的时候常会遇到,对此需要考虑以下情形:
1. node1和node2指向同一节点,这个如何处理
2. node1或node2有不为叶子节点的可能性吗
3. node1或node2一定在树中吗
还要考虑一个效率问题,上述代码中用了两个递归函数,而且存在不必要的递归过程,仔细思考,其实一个递归过程足以解决此问题
实现代码如下:
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
|
#include <iostream> using namespace std; struct Node { Node( int i = 0, Node *pLeft = NULL, Node *pRight = NULL) : data(i), left(pLeft), right(pRight) {} int data; Node *left; Node *right; }; Node *constructNode(Node **pNode1, Node **pNode2) { Node *node12 = new Node(12); Node *node11 = new Node(11); Node *node10 = new Node(10); Node *node9 = new Node(9, NULL, node12); Node *node8 = new Node(8, node11, NULL); Node *node7 = new Node(7); Node *node6 = new Node(6); Node *node5 = new Node(5, node8, node9); Node *node4 = new Node(4, node10); Node *node3 = new Node(3, node6, node7); Node *node2 = new Node(2, node4, node5); Node *node1 = new Node(1, node2, node3); *pNode1 = node6; *pNode2 = node5; return node1; } bool lowestFather(Node *root, Node *node1, Node *node2, Node *&dest) { if (root == NULL || node1 == NULL || node2 == NULL || node1 == node2) return false ; if (root == node1 || root == node2) return true ; bool leftFlag = lowestFather(root->left, node1, node2, dest); bool rightFlag = lowestFather(root->right, node1, node2, dest); if (leftFlag == true && rightFlag == true ) { dest = root; } if (leftFlag == true || rightFlag == true ) return true ; } int main() { Node *node1 = NULL; Node *node2 = NULL; Node *root = constructNode(&node1, &node2); bool flag1 = false ; bool flag2 = false ; Node *dest = NULL; bool flag = lowestFather(root, node1, node2, dest); if (dest != NULL) { cout << "lowest common father: " << dest->data << endl; } else { cout << "no common father!" << endl; } return 0; } |
下面再换一种方式的写法如下:
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
|
#include <iostream> using namespace std; struct Node { Node( int i = 0, Node *pLeft = NULL, Node *pRight = NULL) : data(i), left(pLeft), right(pRight) {} int data; Node *left; Node *right; }; Node *constructNode(Node **pNode1, Node **pNode2) { Node *node12 = new Node(12); Node *node11 = new Node(11); Node *node10 = new Node(10); Node *node9 = new Node(9, NULL, node12); Node *node8 = new Node(8, node11, NULL); Node *node7 = new Node(7); Node *node6 = new Node(6); Node *node5 = new Node(5, node8, node9); Node *node4 = new Node(4, node10); Node *node3 = new Node(3, node6, node7); Node *node2 = new Node(2, node4, node5); Node *node1 = new Node(1, node2, node3); *pNode1 = node11; *pNode2 = node12; return node1; } Node* lowestFather(Node *root, Node *node1, Node *node2) { if (root == NULL || node1 == NULL || node2 == NULL || node1 == node2) return NULL; if (root == node1 || root == node2) return root; Node* leftFlag = lowestFather(root->left, node1, node2); Node* rightFlag = lowestFather(root->right, node1, node2); if (leftFlag == NULL) return rightFlag; else if (rightFlag == NULL) return leftFlag; else return root; } int main() { Node *node1 = NULL; Node *node2 = NULL; Node *root = constructNode(&node1, &node2); bool flag1 = false ; bool flag2 = false ; Node *dest = NULL; Node* flag = lowestFather(root, node1, node2); if (flag != NULL) { cout << "lowest common father: " << flag->data << endl; } else { cout << "no common father!" << endl; } return 0; } |
希望本文所述对大家C++程序算法设计的学习有所帮助。