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);
}