樹的最近公共最先

面試題68-2：二叉樹的最近公共祖先.

class Solution {
public:
	TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
		if (root == NULL || p == NULL || q == NULL) return NULL;
		vector<TreeNode*> path1, path2;   //雙向鏈表容器list也可以
		GetNodePath(root, p, path1);
		GetNodePath(root, q, path2);
		return GetLastCommonNode(path1, path2);
	}
	
    //獲取從根節點到目標節點的路徑
	bool GetNodePath(TreeNode* root, TreeNode* node, vector<TreeNode*>& path){  //此處必須爲引用，因爲此處沒有返回值，改變引用就是改變傳進來的path
        //先把root節點加入到list中
		path.push_back(root);
		//找到目標節點
		if (root == node)  return true;
		//標誌符found
		bool found = false;		//也可以用一個一維標記數組，標記已經訪問過的節點
		if (!found && root->left != NULL) found = GetNodePath(root->left, node, path);
		if (!found && root->right != NULL) found = GetNodePath(root->right, node, path);
		//如果沒有找到，則返回頭節點時，刪除當前節點
		if (!found)
			path.pop_back();
		return found;
	}
	
    //獲取共同節點
	TreeNode* GetLastCommonNode(vector<TreeNode*> path1, vector<TreeNode*> path2){
		TreeNode* pLast = nullptr;
		vector<TreeNode*>::iterator it1 = path1.begin();
		vector<TreeNode*>::iterator it2 = path2.begin();
		while (it1!=path1.end()&&it2!=path2.end()){
			if (*it1 == *it2) pLast = *it1;
			it1++;
			it2++;
		}
		return pLast;
	}
};