劍指offer——樹中兩個節點的最低公共祖先

代碼來源與《劍指offer》

獲得從根節點開始到輸入的兩個結點的兩條，須要遍歷兩次樹，每遍歷一次的時間複雜度是O(n)，獲得的兩條路徑的長度在最差狀況時是O(n)，一般狀況下兩條路徑的長度是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);
}