897. Increasing Order Search Tree ----E

Q:

Given a binary search tree, rearrange the tree in in-order so that the leftmost node in the tree is now the root of the tree, and every node has no left child and only 1 right child.

Example 1:
Input: [5,3,6,2,4,null,8,1,null,null,null,7,9]

       5
      / \
    3    6
   / \    \
  2   4    8
 /        / \ 
1        7   9

Output: [1,null,2,null,3,null,4,null,5,null,6,null,7,null,8,null,9]

 1
  \
   2
    \
     3
      \
       4
        \
         5
          \
           6
            \
             7
              \
               8
                \
                 9  

 

Constraints:

• The number of nodes in the given tree will be between 1 and 100.
• Each node will have a unique integer value from 0 to 1000.

A:

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* increasingBST(TreeNode* root) {
        if(!root)
            return root;
        TreeNode * left = root->left;
        root->left = nullptr;
        TreeNode * newRoot= root;
        if(left){
            newRoot = increasingBST(left);
            TreeNode *tail = newRoot;
            while(tail->right)
                tail = tail->right;
            tail->right = root;
        }
        root->right = increasingBST(root->right);
        return newRoot;
    }
};

NOTE：

上面有个跑到最左边的while loop，这样是很不好的。

因此我改进了下： 每次返回一个最左边和最右边的数组

class Solution {
private:
    vector<TreeNode*> helper(TreeNode* root){
        vector<TreeNode*> res;
        if(!root){
            return res;
        }
        auto resLeft = helper(root->left);
        auto resRight = helper(root->right);
        root->left = NULL;
        if(!resLeft.empty()){
            resLeft.back()->right = root;
            res.push_back(resLeft.front());
        }
        res.push_back(root);
        if(!resRight.empty()){
            root->right = resRight.front();
            res.push_back(resRight.back());
        }
        return res;
    }
public:
    TreeNode* increasingBST(TreeNode* root) {
        vector<TreeNode*> res = helper(root);
        return res.size()?res[0]:NULL;
    }
};

然而，这样效率上已经挺高了。可是依然被很多给beat了。

再改进下：每次更新之前留下的最右边的那个指针（作为参数）。 然后返回最左边的指针。 

注意：  这里，花了我快1个小时，就是C++语法的的锅

改进是： 每次返回最左边的root，但是，有一个跟随的指针的指针（double pointer）

来记录每次的最右边的ancestor (previous) node.  

一开始自己只用了一层指针，是不对的。

class Solution {

private:
    TreeNode* inorder(TreeNode* root, TreeNode** ppre){
        // root is not nullptr
        TreeNode* newRoot = root;
        if(root->left){
            newRoot = inorder(root->left, ppre);
            root->left = nullptr;
        }
        (*ppre)->right = root;
        *ppre = (*ppre)->right;
        if(root->right){
            inorder(root->right, ppre);
        }
        return newRoot;
    }
    
public:
    TreeNode* increasingBST(TreeNode* root) {
        TreeNode* dummy = new TreeNode();
        if(!root)
            return root;
        return inorder(root, &dummy);
    }
};