题解 | 二叉搜索树与双向链表

/*
struct TreeNode {
	int val;
	struct TreeNode *left;
	struct TreeNode *right;
	TreeNode(int x) :
			val(x), left(NULL), right(NULL) {
	}
};*/
class Solution {
public:
    TreeNode* Convert(TreeNode* pRootOfTree) {
        if(!pRootOfTree || (!pRootOfTree->left && !pRootOfTree->right)) {
			return pRootOfTree;
		}

		auto left_head = Convert(pRootOfTree->left);
		auto right_head = Convert(pRootOfTree->right);
		if(left_head) {
			auto left_tail = left_head;
			while(left_tail->right) {
				left_tail = left_tail->right;
			}
			left_tail->right = pRootOfTree;
			pRootOfTree->left = left_tail;
		}

		if(right_head) {
			right_head->left = pRootOfTree;
			pRootOfTree->right = right_head;
		}

		if(left_head) {
			return left_head;
		} else {
			return pRootOfTree;
		}
    }
};

典型递归