C++开发中最小公共父节点的实现方法
来源：爱站网时间：2021-04-02编辑：网友分享
节点通常都要从根节点查找路径，然而从根节点到要查找两个节点的路径，最后得到这两条路径的最后一个交集，爱站技术频道小编来为大家解答C++开发中最小公共父节点的实现方法。
节点通常都要从根节点查找路径，然而从根节点到要查找两个节点的路径，最后得到这两条路径的最后一个交集，爱站技术频道小编来为大家解答C++开发中最小公共父节点的实现方法。
具体方法如下：
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：
#includeusing 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 data data data data
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：
#includeusing 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 data
下面再换一种方式的写法如下：
#includeusing 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 data
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