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)
            return NULL;
        if (!pRootOfTree->left && !pRootOfTree->right)
            return pRootOfTree;
        TreeNode* rootLeft = nullptr, *rootRight = nullptr;
        rootLeft = Convert(pRootOfTree->left);
        rootRight = Convert(pRootOfTree->right);
        if (rootLeft) {
            auto p = rootLeft;
            while (p && p->right) {
                p = p->right;
            }
            p->right = pRootOfTree;
            pRootOfTree->left = p;
        }
        if (rootRight) {
            rootRight->left = pRootOfTree;
            pRootOfTree->right = rootRight;
        }
        return rootLeft ? rootLeft : pRootOfTree;
    }
};