[英]Lowest common ancestor of two leaves in a tree
我有以下樹結構:
struct Node {
int data;
Node* parent = nullptr;
}
每個節點最多有一個父節點,但可以有多個子節點。 我試圖找到沒有任何孩子的兩個節點(節點 1 和節點 2)的最低共同祖先。
這是我當前的代碼:
std::vector<Node*> ancestors1;
std::vector<Node*> ancestors2;
temp_node = node1->parent;
while(temp_node!=nullptr) {
ancestors1.push_back(temp_node);
temp_node = temp_node->parent;
}
temp_node = node2->parent;
while(temp_node!=nullptr) {
ancestors2.push_back(temp_node);
temp_node = temp_node->parent;
}
Node* common_ancestor = nullptr;
if (ancestors1.size() < ancestors2.size()) {
ptrdiff_t t = ancestors1.end() - ancestors1.begin();
std::vector<Node*>::iterator it1 = ancestors1.begin();
std::vector<Node*>::iterator it2 = ancestors2.end() - t;
while(it1!=ancestors1.end()) {
if (*it1 == *it2) {
common_ancestor = *it1;
}
++it1;
}
} else {
ptrdiff_t t = ancestors2.end() - ancestors2.begin();
std::vector<Node*>::iterator it2 = ancestors2.begin();
std::vector<Node*>::iterator it1 = ancestors1.end() - t;
while(it2!=ancestors2.end()) {
if (*it1 == *it2) {
common_ancestor = *it1;
}
++it2;
}
}
return common_ancestor
這段代碼並不總是有效,我不知道為什么。
對不起,我忍不住了。
除了錯別字和錯誤,我相信它看起來更簡單:
#include <cassert>
#include <algorithm>
#include <iostream>
#include <vector>
struct Node {
int data;
Node *parent = nullptr;
};
Node* findCommonAncestor(Node *pNode1, Node *pNode2)
{
// find paths of pNode1 and pNode2
std::vector<Node*> path1, path2;
for (; pNode1; pNode1 = pNode1->parent) path1.push_back(pNode1);
for (; pNode2; pNode2 = pNode2->parent) path2.push_back(pNode2);
// revert paths to make indexing simple
std::reverse(path1.begin(), path1.end());
std::reverse(path2.begin(), path2.end());
// compare paths
Node *pNode = nullptr; // ancestor
size_t n = std::min(path1.size(), path2.size());
for (size_t i = 0; i < n; ++i) {
if (path1[i] == path2[i]) pNode = path1[i];
else break;
}
// done
return pNode;
}
int main()
{
// sample tree:
/* 1
* |
* 2
* / \
* 3 4
* |
* 5
*/
Node node1 = { 1, nullptr };
Node node2 = { 2, &node1 };
Node node3 = { 3, &node2 };
Node node4 = { 4, &node2 };
Node node5 = { 5, &node4 };
Node *pNode = findCommonAncestor(&node3, &node5);
if (pNode) {
std::cout << "Lowest common ancestor: " << pNode->data << '\n';
} else {
std::cout << "No common ancestor found!\n";
}
}
輸出:
Lowest common ancestor: 2
筆記:
雖然使用iterator
有助於保持代碼的通用性……
我認為這是堅持使用普通舊數組(又名std::vector
)索引可以簡化事情的情況之一。
我發現了問題。 除了需要移動兩個迭代器而不是一個(感謝 Jarod42 和 v78),我還需要在找到最低的公共祖先時立即跳出 while 循環(否則它返回最高的公共祖先)。
while(it1!=ancestors1.end()) {
if (*it1 == *it2) {
common_ancestor = *it1;
break;
}
你應該不需要任何額外的空間來解決這個問題:
// measure depths
size_t depth1=0;
for (Node *n = node1; n; n=n->parent, ++depth1);
size_t depth2=0;
for (Node *n = node2; n; n=n->parent, ++depth2);
// move the deeper one up until they're the same depth
for (;depth1 > depth2; node1 = node1->parent, --depth1);
for (;depth2 > depth1; node2 = node2->parent, --depth2);
// move them both up until they match
while(node1 != node2) {
node1 = node1->parent;
node2 = node2->parent;
}
return node1;
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