Question

我正在尝试获取树中两个节点的最不常见的祖先。我试过了，但问题是if one node is the descendant node for other我无法获得LCA。
我尝试解决它，然后它只适用于后代节点。不知道如何处理这个问题。

Node* Tree::LCA(Node* root, Node* n1, Node* n2) {
    list<Node*> a1,a2;

    while(n1 != NULL) {
        a1.push_back(n1->parent);
        n1 = n1->parent;
    }

    while(n2 != NULL) {
        a2.push_back(n2->parent);
        n2 = n2->parent;
    }

    while(!a1.empty() && !a2.empty() && a1.back() == a2.back()) {   
        a1.pop_back();
        a2.pop_back();
    }

    if( a1.back() != a2.back()) {
        Node* rn = a1.back();
        cout << " LCA of r-U and r_v is " << rn->index << endl;
    }
}

Answer 1

您开始从n1->parent和n2->parent推送。而是在推送他们的父母和其他祖先之前推送n1和n2。所以你的代码应该是：

Node* Tree::LCA(Node* root, Node* n1, Node* n2) {
    list<Node*> a1,a2;
    a1.push_back(n1); // line to be added

    while(n1 != NULL) {
        a1.push_back(n1->parent);
        n1 = n1->parent;
    }

    a2.push_back(n2); // line to be added
    while(n2 != NULL) {
        a2.push_back(n2->parent);
        n2 = n2->parent;
    }
    // rest of code

Answer 2

Node* Tree::LCA(Node* root, Node* n1, Node* n2) {
    list<Node*> a1,a2;

    while(n1 != NULL) {
        a1.push_back(n1); // push n1 
        n1 = n1->parent;
    }

    while(n2 != NULL) {
        a2.push_back(n2);  // push n2
        n2 = n2->parent;
    }

    Node* old; // create new node
    while(!a1.empty() && !a2.empty() && a1.back() == a2.back()) {
        old = a1.back(); // store the node before popping
        a1.pop_back();
        a2.pop_back();
    }

    if( a1.back() != a2.back()) {
       // Node* rn = a1.back();  //not needed
        cout << " LCA of r-U and r_v is " << old->index << endl; // changed 
    }
}

这可能会有所帮助。

LCA后代节点

2 个答案: