二叉树求两节点最低公共祖先，求随意两节点距离

-------1.求最低公共祖先LCA( Lowest Common Ancestor ）

         什么是最低公共祖先？例如以下图。2与3的LCA是1；1与4的LCA是1；4与5的LCA是2。

      那么给定两个节点n1和n2，如今要找出它们的LCA，怎样找？首先分析一下，n1和n2有几种情况？第一种。n1和n2分别在一个节点的左右子树中，比方4和5，LCA就是2，5和6，LCA就是1。2和3，LCA就是1；另外一种。n1和n2都在左子树或右子树，比方2和4，LCA就是2，1和4。LCA就是1。

一共这两种情况，事实上一点也不难。看代码实现吧。

/* 仅仅用一次遍历解决LCA */
#include <iostream>
using namespace std;

struct Node
{
    Node *left, *right;
	int key;
};

Node* newNode(int key)
{
	Node *temp = new Node;
	temp->key = key;
	temp->left = temp->right = nullptr;
	return temp;
}

// 返回n1和n2的LCA的指针
Node * findLCA(Node* root, int n1, int n2)
{
	if (root == nullptr)
		return nullptr;

	if (root->key == n1 || root->key == n2)
		return root;

	Node * left = findLCA(root->left, n1, n2);
	Node * right = findLCA(root->right, n1, n2);

	if (left&&right)//假设n1和n2分别在左右子树中
		return root;

	return (left != nullptr ?
 left : right);
}

//測试
int main()
{
	// 构造上面图中的树
	Node * root = newNode(1);
	root->left = newNode(2);
	root->right = newNode(3);
	root->left->left = newNode(4);
	root->left->right = newNode(5);
	root->right->left = newNode(6);
	root->right->right = newNode(7);
	cout << "LCA(4, 5) = " << findLCA(root, 4, 5)->key;
	cout << "\nLCA(4, 6) = " << findLCA(root, 4, 6)->key;
	cout << "\nLCA(3, 4) = " << findLCA(root, 3, 4)->key;
	cout << "\nLCA(2, 4) = " << findLCA(root, 2, 4)->key;
	cout << endl;
	return 0;
}

        对于普通的二叉树，怎样找到两个给定节点之家的距离？距离是指连接两个节点须要的最小的边的条数。

比如以下的二叉树：

代码例如以下：

#include <iostream>
using namespace std;

struct Node
{
    Node *left, *right;
	int key;
};

Node* newNode(int key)
{
	Node *temp = new Node;
	temp->key = key;
	temp->left = temp->right = nullptr;
	return temp;
}

int FindLevel(Node * _root, int node)//返回node节点在root中的第几层。-1代表在子树下未找到node
{
	if (_root == nullptr)
		return -1;

	if (_root->key == node)
		return 0;

	int level = FindLevel(_root->left, node);//先在左子树找

	if (level == -1)//假设左子树没找到，在右子树找
		level = FindLevel(_root->right, node);

	if (level != -1)//在找到node的时候，递归一个一个结束，逐渐加一
		return level + 1;

	return -1;
}

Node * findLCA(Node* root, int n1, int n2)//找到n1和n2的LCA（已在上部分讲过怎样求解LCA的问题）
{
	if (root == nullptr)
		return nullptr;

	if (root->key == n1 || root->key == n2)
		return root;

	Node * left = findLCA(root->left, n1, n2);
	Node * right = findLCA(root->right, n1, n2);

	if (left&&right)//假设n1和n2分别在左右子树中
		return root;

	return (left != nullptr ?
 left : right);

}

int DistanceNodes(Node * _root, int n1, int n2)
{
	Node * lca = findLCA(_root, n1, n2);

	int lca_level = FindLevel(_root, lca->key);
	int n1_level = FindLevel(_root, n1);
	int n2_level = FindLevel(_root, n2);

	return (n1_level - lca_level) + (n2_level - lca_level);//也就是n1_level+n2_level-2*lca_level，只是前者这样写更easy理解
}

int main()
{
	Node * root = newNode(1);
	root->left = newNode(2);
	root->right = newNode(3);
	root->left->left = newNode(4);
	root->left->right = newNode(5);
	root->right->left = newNode(6);
	root->right->right = newNode(7);
	root->right->left->right = newNode(8);

	cout << "Dist(8,7) = " << DistanceNodes(root, 8, 7) << endl;
	cout << "Dist(8,3) = " << DistanceNodes(root, 8, 3) << endl;
	cout << "Dist(8,2) = " << DistanceNodes(root, 8, 2) << endl;

	return 0;
}

二叉树----第一篇（构造，遍历。求节点，求叶子）參考链接：http://blog.csdn.net/laojiu_/article/details/50196823

參考自：http://www.acmerblog.com/distance-between-given-keys-5995.html