求二叉树中两个节点的最近公共祖先结点

二叉树是搜索二叉树

1、原理：二叉搜索树是排序过的 ，位于左子树的结点都比父结点小，位于右子树的结点都比父结点大，我们只需从根节点开始和两个输入的结点进行比较，如果当前节点的值比两个结点的值都大，那么最低的公共祖先结点一定在该结点的左子树中，下一步开遍历当前结点的左子树。如果当前节点的值比两个结点的值都小，那么最低的公共祖先结点一定在该结点的右子树中，下一步开遍历当前结点的右子树。这样从上到下找到第一个在两个输入结点的值之间的结点。

2、实现代码

BinaryNode* CreateBinaryTree(int* array, int length){BinaryNode* root = new BinaryNode(array[0]);BinaryNode* pTemp = NULL;for (int idx = 1; idx < length; idx++){BinaryNode* pCur = root;//寻找插入位置while (pCur){if (array[idx]<pCur->_value){pTemp = pCur;pCur = pCur->_left;}else if (array[idx]>pCur->_value){pTemp = pCur;pCur = pCur->_right;}elsereturn NULL;}//插入元素pCur = new BinaryNode(array[idx]);if (array[idx] < pTemp->_value)pTemp->_left = pCur;elsepTemp->_right = pCur;}return root;}//搜素二叉树寻找最近公共祖先节点BinaryNode* AncestorTreeNode(BinaryNode* pRoot, BinaryNode* node1, BinaryNode* node2){if (pRoot == NULL)return NULL;if (node1 == NULL)return node2;if (node2 == NULL)return node1;while (pRoot){if (pRoot->_value > node1->_value && pRoot->_value > node2->_value)pRoot = pRoot->_left;if (pRoot->_value < node1->_value && pRoot->_value < node2->_value)pRoot = pRoot->_right;elsereturn pRoot;}return NULL;}

测试用例：

void FunTest1(){BinaryNode* pRoot = NULL;int array[] = { 5, 2, 6 };int length = sizeof(array) / sizeof(array[0]);pRoot = CreateBinaryTree(array, length);BinaryNode* node1 = pRoot->_left->_left;BinaryNode* node2 = pRoot->_left->_right;    BinaryNode* pNode = AncestorTreeNode(pRoot, node1, node2);if (pNode)cout << pNode->_value << endl;}

二叉树节点中包含指向父节点的指针

1、原理：如果树中每个结点都有父结点（根结点除外），这个问题可以转换成求两个链表的第一个公共结点，假设树结点中指向父结点的指针是parent，树的每一个叶结点开始都由一个指针parent串起来的链表，每个链表的尾结点就是树的根结点。那么输入的这两个结点位于链表上，它们的最低公共祖先结点刚好是这两个链表的第一个公共结点。

2、实现代码

int Hight(BinaryNode* root, BinaryNode* node){int len = 0;while (node){len++;node = node->_parent;}return len;}BinaryNode* GetLastCommonAncestor(BinaryNode* root, BinaryNode* node1, BinaryNode* node2){if (root == NULL || node1 == NULL || node2 == NULL)return NULL;int len1 = Hight(root, node1);//一个链表的高度int len2 = Hight(root, node2);//另一个链表的高度//寻找两个链表的第一个交点while (len1 != len2){if (len1 < len2)len2--;elselen1--;}while (node1 && node2 && node1 != node2){node1 = node1->_parent;node2 = node2->_parent;}if (node1 == node2)return node1;elsereturn NULL;}

测试用例

void FunTest(){BinaryNode* root = new BinaryNode(1);BinaryNode* cur = root;queue<BinaryNode*> q;BinaryNode* top = NULL;q.push(root);for (int i = 2; i <= 7; i++){if (!q.empty()){top = q.front();if (cur == top->_left){cur = new BinaryNode(i);top->_right = cur;cur->_parent = top;q.pop();}else{cur = new BinaryNode(i);top->_left = cur;cur->_parent = top;}q.push(cur);}}BinaryNode* node1 = root->_left->_left;BinaryNode* node2 = root->_left->_right;BinaryNode* ancestor = GetLastCommonAncestor(root, node1, node2);if (ancestor)cout << ancestor->_value << endl;elsecout << "没有公共祖先" << endl;}

二叉树是普通二叉树， 没有指向父结点的指针

1、原理：在二叉树根结点 的左子树和右子树中分别找输入的两个结点，如果两个结点都在左子树，遍历当前结点的左子树，如果两个结点都在右子树，遍历当前结点的右子树，直到一个在当前结点的左子树，一个在当前结点的右子树，返回当前结点就是最低的公共祖先结点。

2、实现代码

bool IsNodeInTree(BinaryNode* pRoot, BinaryNode* pNode){if (pRoot == NULL || pNode == NULL)return false;if (pNode == pRoot)return true;if (IsNodeInTree(pRoot->_left, pNode) || IsNodeInTree(pRoot->_right, pNode))return true;return false;}BinaryNode* GetCommonAncestor(BinaryNode* pRoot, BinaryNode* node1, BinaryNode* node2){if (pRoot == NULL)return NULL;if (node1 == pRoot && IsNodeInTree(pRoot, node2) || node2 == pRoot && IsNodeInTree(pRoot, node1))return pRoot;bool node1left = IsNodeInTree(pRoot->_left, node1);bool node1right = IsNodeInTree(pRoot->_right, node1);bool node2left = IsNodeInTree(pRoot->_left, node2);bool node2right = IsNodeInTree(pRoot->_right, node2);if (node1left && node2right || node1right && node2left)return pRoot;if (node1left && node2left)return GetCommonAncestor(pRoot->_left, node1, node2);if (node1right && node2right)return GetCommonAncestor(pRoot->_right, node1, node2);return NULL;}

测试用例

void FunTest3(){BinaryNode* root = new BinaryNode(1);BinaryNode* cur = root;queue<BinaryNode*> q;BinaryNode* top = NULL;q.push(root);for (int i = 2; i <= 7; i++){if (!q.empty()){top = q.front();if (cur == top->_left){cur = new BinaryNode(i);top->_right = cur;cur->_parent = top;q.pop();}else{cur = new BinaryNode(i);top->_left = cur;cur->_parent = top;}q.push(cur);}}BinaryNode* node1 = root->_left->_left;BinaryNode* node2 = root->_left->_right;BinaryNode* ancestor = GetCommonAncestor(root, node1, node2);if (ancestor)cout << ancestor->_value << endl;elsecout << "没有公共祖先" << endl;}

头文件和结点的结构体

#include<iostream>#include<queue>using namespace std;struct BinaryNode{BinaryNode* _left;BinaryNode* _right;BinaryNode* _parent;int _value;BinaryNode(const int& value):_value(value), _left(NULL), _right(NULL), _parent(NULL){}};