二叉搜索树与双向链表

题目描述

代码

class Solution {
public:
TreeNode* Convert(TreeNode* pRootOfTree)
{
TreeNode* lastNode = NULL;
convert(pRootOfTree, &lastNode);
while (lastNode != NULL && lastNode->left != NULL) {
lastNode = lastNode->left;
};
return lastNode;
}

void convert(TreeNode* root, TreeNode** lastNode)//中序遍历就是排序后的结果
{
if (root == NULL) {
return;
}

//左节点
if (root->left != NULL) {
convert(root->left, lastNode);
}
root->left = *lastNode;
if (*lastNode != NULL) {
(*lastNode)->right = root;
}

*lastNode = root;

//右节点
if (root->right != NULL) {
convert(root->right, lastNode);
}
}
};