【数据结构】求二叉树中两个结点最近的公共祖先

港控/mmm° 2022-06-11 03:59 151阅读 0赞

对于两个结点最近的公共祖先，有三种情况：

1，该二叉树是一棵二叉搜索树

struct TreeNode
    {
    	int _value;
    	TreeNode* _left;
    	TreeNode* _right;
    
    	TreeNode(int x)
    		: _value(x)
    		, _left(NULL)
    		, _right(NULL)
    	{}
    };
    TreeNode* LowestAncestor(TreeNode* _root, TreeNode* node1, TreeNode* node2)
    {
    	int m1 = min(node1->_value, node2->_value);
    	int m2 = max(node1->_value, node2->_value);
    	while (true)
    	{
    		if (_root->_value > m2)
    		{
    			_root = _root->_left;
    		}
    		else if (_root->_value < m1)
    		{
    			_root = _root->_right;
    		}
    		else
    		{
    			break;
    		}
    	}
    	return _root;
    }

2，该二叉树是一棵具有三叉链的二叉树

struct TreeNode
    {
    	int _value;
    	TreeNode* _left;
    	TreeNode* _right;
    	TreeNode* _parent;
    
    	TreeNode(int x)
    		:_value(x)
    		, _left(NULL)
    		, _right(NULL)
    		, _parent(NULL)
    	{}
    };
    
    TreeNode* LowestAncestor(TreeNode* _root, TreeNode* node1, TreeNode* node2)
    {
    	TreeNode* cur = NULL;
    	while (node1 != NULL)
    	{
    		node1 = node1->_parent;
    		cur = node2;
    		while (cur != NULL)
    		{
    			if (node1 == cur->_parent)
    			{
    				return node1;
    			}
    			cur = cur->_parent;
    		}
    	}
    }

3，一般的普通二叉树：

struct TreeNode
    {
    	int _value;
    	TreeNode* _left;
    	TreeNode* _right;
    
    	TreeNode(int x)
    		:_value(x)
    		, _left(NULL)
    		, _right(NULL)
    	{}
    };
    
    TreeNode* LowestAncestor(TreeNode* _root, TreeNode* node1, TreeNode* node2)
    {
    	if (_root == NULL || _root == node1 || _root == node2)
    	{
    		return _root;
    	}
    	TreeNode* left = LowestAncestor(_root->_left, node1, node2);
    	TreeNode* right = LowestAncestor(_root->_right, node1, node2);
    
    	if (left != NULL&&right = !NULL)
    	{
    		return _root;
    	}
    	return left = NULL ? right : left;
    }