本文實例講述了C++實現尋找最低公共父節點的方法，是數據結構中二叉樹的經典算法。分享給大家供大家參考。具體方法如下：
最低公共父節點，意思很好理解。
思路1：最低公共父節點滿足這樣的條件：兩個節點分別位于其左子樹和右子樹，那么定義兩個bool變量，leftFlag和rightFlag，如果在左子樹中，leftFlag為true，如果在右子樹中，rightFlag為true，僅當leftFlag == rightFlag == true時，才能滿足條件。
實現代碼如下：
#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; }}這類問題在面試的時候常會遇到，對此需要考慮以下情形：
1. node1和node2指向同一節點，這個如何處理
2. node1或node2有不為葉子節點的可能性嗎
3. node1或node2一定在樹中嗎
還要考慮一個效率問題，上述代碼中用了兩個遞歸函數，而且存在不必要的遞歸過程，仔細思考，其實一個遞歸過程足以解決此問題
實現代碼如下：
#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;}下面再換一種方式的寫法如下：
#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;}希望本文所述對大家C++程序算法設計的學習有所幫助。
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