剑指offer——树中两个节点的最低公共祖先

代码来源与《剑指offer》

得到从根节点开始到输入的两个结点的两条，需要遍历两次树，每遍历一次的时间复杂度是O(n)，得到的两条路径的长度在最差情况时是O(n)，通常情况下两条路径的长度是O(logn)。

#include <iostream>#include <vector>#include <list>using namespace std;struct TreeNode {    int m_nValue;        std::vector<TreeNode*> m_vChildren;   };bool GetNodePath(TreeNode *pRoot,TreeNode *pNode, list<TreeNode*> &path){if (pRoot=pNode){return true;}path.push_back(pRoot);bool found=false;vector<TreeNode*>::iterator i=pRoot->m_vChildren.begin();while(!found&&i<pRoot->m_vChildren.end()){found=GetNodePath(*i,pNode,path);++i;}if (!found){path.pop_back();}return found;}TreeNode* GetLastCommonNode(const list<TreeNode*>& path1, const list<TreeNode*>& path2){list<TreeNode*>::const_iterator i1=path1.begin();list<TreeNode*>::const_iterator i2=path2.begin();TreeNode* pLast=NULL;while(i1!=path1.end()&&i2!=path2.end()){if (*i1==*i2){pLast=*iterator;}i1++;i2++;}return pLast;}TreeNode* GetLastCommonParent(TreeNode* pRoot,TreeNode* pNode1,TreeNode* pNode2){if (pRoot==NULL||pNode1==NULL||pNode2==NULL){return NULL;}list<TreeNode*> path1;GetNodePath(pRoot,pNode1,path1);list<TreeNode*> path2;GetNodePath(pRoot,pNode2,path2);return GetLastCommonNode(path1,path2);}