C++实现寻找最低公共父节点的方法
时间：2021-01-03 15:36:30|栏目：C代码|点击： 次
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：
#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：
#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：
#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。
