二叉树中两个节点的最近公共父节点

来源：互联网发布：数据库发展趋势编辑：程序博客网时间：2024/05/22 00:12

情况一：root未知，但是每个节点都有parent指针

解决方法是求出linkedList A的长度lengthA, linkedList B的长度LengthB。然后让长的那个链表走过abs(lengthA-lengthB)步之后，齐头并进，相遇的节点就是最近公共父节点。

template <class T>int getLength(BinaryTreeNode<T>* pNode){      int length = 0;BinaryTreeNode<T>* pTemp = pNode;while (pTemp){length++ ;   pTemp = pTemp->pParent; }return length;    }template <class T>BinaryTreeNode<T>* findLCACase1(BinaryTreeNode<T>* pNode1, BinaryTreeNode<T>* pNode2){int length1 = getLength(pNode1);int length2 = getLength(pNode2);// skip the abs(length1-length2)BinaryTreeNode<T>* pIter1 = NULL;BinaryTreeNode<T>* pIter2 = NULL;int k = 0;   if (length1>=length2) {   bstNode* pTemp = pNode1;while (k++ < length1-length2) {pTemp = pTemp->pParent;    }pIter1 = pTemp;pIter2 = pNode2;   } else {BinaryTreeNode* pTemp = pNode1;   while (k++ < length2-length1) {pTemp = pTemp->pParent;    }pIter1 = pNode1;pIter2 = pTemp;   }while (pIter1 && pIter2 && pIter1!= pIter2) {pIter1 = pIter1->pParent;pIter2 = pIter2->pParent;   }return pIter1;   }

情况二：节点只有left/right，没有parent指针，root已知

template <class T>bool nodePath(BinaryTreeNode<T>* pRoot, int value, vector<BinaryTreeNode<T>*>& path)   {   if (pRoot==NULL)return false;if (pRoot->data != value) {if (nodePath(pRoot->pLeft, value, path)) {path.push_back(pRoot);return true;   } else if (nodePath(pRoot->pRight, value, path)) {path.push_back(pRoot);return true;   } elsereturn false;      } else {path.push_back(pRoot);return true;   } }template <class T>BinaryTreeNode<T>* findLCACase2(BinaryTreeNode<T>* pNode, int value1, int value2)   {std::vector<BinaryTreeNode<T>*> path1;std::vector<BinaryTreeNode<T>*> path2;bool find = false;find |= nodePath(pNode, value1, path1);find &= nodePath(pNode, value2, path2);BinaryTreeNode<T>* pReturn = NULL;if (find) {int minSize = path1.size()>path2.size()?path2.size():path1.size();int it1 = path1.size()- minSize;int it2 = path2.size()- minSize;for ( ; it1<path1.size(), it2<path2.size(); it1++, it2++) {if (path1[it1]==path2[it2]) {pReturn = path1[it1];break;   }}}return pReturn;   }

情况三：二叉树是个二叉查找树，且root和两个节点的值(a, b)已知

BinaryTreeNode<T>* findLCACase3(BinaryTreeNode<T>* pNode, int value1, int value2)   {BinaryTreeNode<T>* pTemp = pNode;while (pTemp) {if (pTemp->data>value1 && pTemp->data>value2)pTemp = pTemp->pLeft;else if(pTemp->data<value1 && pTemp->data<value2)pTemp = pTemp->pRight;elsereturn pTemp;   }return NULL;   }