acwing 49. 二叉搜索树与双向链表

地址:https://www.acwing.com/problem/content/87/

输入一棵二叉搜索树，将该二叉搜索树转换成一个排序的双向链表。

要求不能创建任何新的结点，只能调整树中结点指针的指向。

注意：

• 需要返回双向链表最左侧的节点。

例如，输入下图中左边的二叉搜索树，则输出右边的排序双向链表。

解法

树的处理 一半都是递归  分为 根  树的左子树 和树的右子树

子树也是一棵树 进行递归处理  向上返回一个双链表  返回链表的头尾

最后全部转化链表

代码

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
 
    TreeNode* rethead = NULL;
 
    TreeNode* gleft = NULL;
    TreeNode* gright = NULL;
 
    void convertInner(TreeNode* root)
    {
        if (NULL == root) return;
 
        if (root->val < rethead->val) rethead = root;
 
        if (root->left == NULL && root->right == NULL) {
            gleft = root; gright = root;
            return;
        }
        else if (root->left != NULL && root->right == NULL) {
            convertInner(root->left);
            gright->right = root;
            root->left = gright;
            gright = root;
        }
        else if (root->right != NULL && root->left == NULL) {
            convertInner(root->right);
            gleft->left = root;
            root->right = gleft;
            gleft = root;
        }
        else if (root->right != NULL && root->left != NULL) {
            convertInner(root->left);
            gright->right = root;
            root->left = gright;
 
            TreeNode* leftcopy = gleft;
 
            convertInner(root->right);
            gleft->left = root;
            root->right = gleft;
            gleft = leftcopy;
        }
    }
 
    TreeNode* convert(TreeNode* root) {
        if (NULL == root) return NULL;
        rethead = root;
 
        if (root->left == NULL && root->right == NULL) return root;
 
        if (root->left != NULL && root->right == NULL) {
            convertInner(root->left);
            gright->right = root;
            root->left = gright;
        }
        else if (root->right != NULL && root->left == NULL) {
            convertInner(root->right);
            gleft->left = root;
            root->right = gleft;
        }
        else if (root->right != NULL && root->left != NULL) {
            convertInner(root->left);
            gright->right = root;
            root->left = gright;
 
            convertInner(root->right);
            gleft->left = root;
            root->right = gleft;
        }
 
        return rethead;
    }
 
};

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
 
    TreeNode* rethead = NULL;
 
    TreeNode* gleft = NULL;
    TreeNode* gright = NULL;
 
    void convertInner(TreeNode* root)
    {
        if (NULL == root) return;
 
        if (root->val < rethead->val) rethead = root;
 
        if (root->left == NULL && root->right == NULL) {
            gleft = root; gright = root;
            return;
        }
        else if (root->left != NULL && root->right == NULL) {
            convertInner(root->left);
            gright->right = root;
            root->left = gright;
            gright = root;
        }
        else if (root->right != NULL && root->left == NULL) {
            convertInner(root->right);
            gleft->left = root;
            root->right = gleft;
            gleft = root;
        }
        else if (root->right != NULL && root->left != NULL) {
            convertInner(root->left);
            gright->right = root;
            root->left = gright;
 
            TreeNode* leftcopy = gleft;
 
            convertInner(root->right);
            gleft->left = root;
            root->right = gleft;
            gleft = leftcopy;
        }
    }
 
    TreeNode* convert(TreeNode* root) {
        if (NULL == root) return NULL;
        rethead = root;
 
        if (root->left == NULL && root->right == NULL) return root;
 
        if (root->left != NULL && root->right == NULL) {
            convertInner(root->left);
            gright->right = root;
            root->left = gright;
        }
        else if (root->right != NULL && root->left == NULL) {
            convertInner(root->right);
            gleft->left = root;
            root->right = gleft;
        }
        else if (root->right != NULL && root->left != NULL) {
            convertInner(root->left);
            gright->right = root;
            root->left = gright;
 
            convertInner(root->right);
            gleft->left = root;
            root->right = gleft;
        }
 
        return rethead;
    }
 
};