Geeks 一般二叉树的LCA

不是BST，那么搜索两节点的LCA就复杂点了，由于节点是无序的。

以下是两种方法，都写进一个类里面了。

当然须要反复搜索的时候。能够使用多种方法加速搜索。

#include <iostream>
#include <vector>
using namespace std;

class LCANormalTree
{
	struct Node
	{
		int key;
		Node *left, *right;
		Node(int k) : key(k) , left(NULL), right(NULL) {}
	};

	//method 1:
	bool findPath(Node *root, vector<Node *> &path, int k)
	{
		if (!root) return false;
		path.push_back(root);
		if (root->key == k) return true;
		if (findPath(root->left, path, k) ||	findPath(root->right, path, k))
			return true;

		path.pop_back();
		return false;
	}

	Node *findLCA(Node *root, int n1, int n2)
	{
		vector<Node *> pathN1, pathN2;
		if (!findPath(root, pathN1, n1) || !findPath(root, pathN2, n2))
			return NULL;

		int i = 0;
		for (; i < (int) pathN1.size() && pathN1[i] == pathN2[i]; i++);
		return pathN1[i-1];
	}

	//Method 2:
	Node *findLCAUtil(Node *root, int n1, int n2, bool &v1, bool &v2)
	{
		if (!root) return NULL;
		if (root->key == n1)
		{
			v1 = true;
			return root;
		}
		if (root->key == n2)
		{
			v2 = true;
			return root;
		}
		Node *left = findLCAUtil(root->left, n1, n2, v1, v2);
		Node *right = findLCAUtil(root->right, n1, n2, v1, v2);

		if (left && right) return root;
		return left?

left:right;
	}

	bool find(Node *root, int k)
	{
		if (!root) return false;
		if (root->key == k || find(root->left, k) || find(root->right, k))
			return true;
		return false;
	}

	Node *findLCA_2(Node *root, int n1, int n2)
	{
		bool v1 = false, v2 = false;
		Node *lca = findLCAUtil(root, n1, n2, v1, v2);
		//当两者在不同一边的时候v1&&v1，当一个为另外一个的单亲节点的时候，那么就出现后面两种情况了。
		if (v1 && v2 || v1 && find(lca, n2) || v2 && find(lca, n1))
			return lca;
		return NULL;
	}

	Node *root;
public:
	LCANormalTree()
	{
		cout<<"Algorithm 1 run\n";
		run_1();
		cout<<"\nAlgorithm 2 run\n";
		run_2();
	}

	void run_1()
	{
		root = new Node(1);
		root->left = new Node(2);
		root->right = new Node(3);
		root->left->left = new Node(4);
		root->left->right = new Node(5);
		root->right->left = new Node(6);
		root->right->right = new Node(7);
		Node *t = findLCA(root, 4, 5);
		cout << "LCA(4, 5) = " << (t? t->key : -1);
		t = findLCA(root, 4, 6);
		cout << "\nLCA(4, 6) = " << (t?

t->key : -1);
		t = findLCA(root, 3, 4);
		cout << "\nLCA(3, 4) = " << (t? t->key : -1);
		t = findLCA(root, 2, 4);
		cout << "\nLCA(2, 4) = " << (t? t->key : -1);
		cout << endl;

		deleteTree(root);
	}

	void run_2()
	{
		root = new Node(1);
		root->left = new Node(2);
		root->right = new Node(3);
		root->left->left = new Node(4);
		root->left->right = new Node(5);
		root->right->left = new Node(6);
		root->right->right = new Node(7);
		Node *lca =  findLCA(root, 4, 5);
		if (lca != NULL)
			cout << "LCA(4, 5) = " << lca->key;
		else
			cout << "Keys are not present ";

		lca =  findLCA(root, 4, 10);
		if (lca != NULL)
			cout << "\nLCA(4, 10) = " << lca->key;
		else
			cout << "\nKeys are not present ";

		cout<<endl;
		deleteTree(root);
	}

	~LCANormalTree()
	{
		if (root) deleteTree(root);
	}

	void deleteTree(Node *&r)
	{//注意不能是*r。要*&r，由于须要改动r为NULL，防止多次释放
		if (r)
		{
			deleteTree(r->left);
			deleteTree(r->right);
			delete r; r = NULL;
		}
	}
};