在二叉树中找最近公共父节点。分为两种情况，一种是有父指针，一种没有父指针。
#include<iostream>
struct Node
{
int data;
Node* left;
Node* right;
Node* parent;
Node() :left(NULL), right(NULL), parent(NULL)
{}
};
int getDpeth(Node *n)//结点n到根节点深度
{
int count = 0;
while (n)
{
++count;
n = n->parent;
}
return count;
}
Node* findNearestCommonAncestor(Node* n1, Node* n2)
{
int depth1 = getDpeth(n1);
int depth2 = getDpeth(n2);
//移动同一深度
while (depth1 > depth2)
{
n1 = n1->parent;
--depth1;
}
while (depth1 < depth2)
{
n2 = n2->parent;
--depth2;
}
//向上找
while (n1 != n2)
{
n1 = n1->parent;
n2 = n2->parent;
}
return n1;
}
int main()
{
//测试
Node* A[11];
for (int i = 0; i < 11; ++i)
{
A[i] = new Node();
A[i]->data = i;
}
for (int i = 0; i < 5; ++i)
{
A[i]->left = A[i * 2 + 1];
A[i * 2 + 1]->parent = A[i];
A[i]->right = A[i * 2 + 2];
A[i * 2 + 2]->parent = A[i];
}
Node* Ancestor = findNearestCommonAncestor(A[7], A[6]);
}
#include<iostream>
struct Node
{
int data;
Node* left;
Node* right;
Node() :left(NULL), right(NULL)
{}
};
//计算当前结点包含n1、n2个数
int countMatch(Node *current, Node* n1, Node* n2)
{
if (current == NULL)
return 0;
int count = countMatch(current->left, n1, n2) + countMatch(current->right, n1, n2);
if (current == n1 || current == n2)
return 1 + count;
return count;
}
Node* findLCA(Node* root, Node* n1, Node* n2)
{
if (root == NULL)
return NULL;
if (root == n1 || root == n2)
return root;
int count = countMatch(root->left, n1, n2);//左子树包含n1和n2的个数
if (count == 1)
return root;//左子树一个，右子树肯定也有一个
else if (count == 2)//都在左子树
return findLCA(root->left, n1, n2);
else//都在右子树
return findLCA(root->right, n1, n2);
}
int main()
{
//测试
Node* A[11];
for (int i = 0; i < 11; ++i)
{
A[i] = new Node();
A[i]->data = i;
}
for (int i = 0; i < 5; ++i)
{
A[i]->left = A[i * 2 + 1];
A[i]->right = A[i * 2 + 2];
}
Node* Ancestor = findLCA(A[0],A[7], A[10]);
}#include<iostream>
struct Node
{
int data;
Node* left;
Node* right;
Node() :left(NULL), right(NULL)
{}
};
Node* findLCA(Node *root, Node* n1, Node* n2)
{
if (root == NULL)//没找到
return NULL;
if (root == n1 || root == n2)//找到
return root;
Node* L = findLCA(root->left, n1, n2);//左子树
Node* R = findLCA(root->right, n1, n2);//右子树
//当前结点左右子树都找到了n1和n2，那么这个结点就是LCA结点
if (L != NULL&R != NULL)
return root;
//否则是不为NULL的结点，或者两个都为NULL
else
return L !=NULL ? L : R;
}
int main()
{
//测试
Node* A[11];
for (int i = 0; i < 11; ++i)
{
A[i] = new Node();
A[i]->data = i;
}
for (int i = 0; i < 5; ++i)
{
A[i]->left = A[i * 2 + 1];
A[i]->right = A[i * 2 + 2];
}
Node* Ancestor = findLCA(A[0], A[7], A[10]);
}原文地址：http://blog.csdn.net/kangroger/article/details/40392925
