今天上午看了一道题，求二叉树的两个节点的最近公共父节点 充分利用递归的特性，写的很好，当时就纠结于不知道如何用递归来存储访问正确的路径，看人家写的，smart

充分利用递归的特性，写的很好，当时就纠结于不知道如何用递归来写，看人家写的，smart

以下代码转自该网站：http://blog.csdn.net/hhygcy/article/details/4660362

#include <vector> bool nodePath (bstNode* pRoot, int value, std::vector<bstNode*>& 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; } else return false; } } else { path.push_back(pRoot); return true; } } bstNode* findLCACase1(bstNode* pNode, int value1, int value2) { std::vector<bstNode*> path1; std::vector<bstNode*> path2; bool find = false; find |= nodePath(pNode, value1, path1); find &= nodePath(pNode, value2, path2); bstNode* 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; }