[LeetCode] 897. Increasing Order Search Tree

Description

Given a 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  

Note:

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

---

Analyse

用二叉树中序遍历的结果生成新的二叉树，中序遍历很容易就写出来了，但生成新的二叉树我卡了好久

问题出在递归和递归函数的传参上

以下面这个二叉树为input

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

第一个要被插入树的节点是1，这个1是在第四层的递归中返回的，将1插入二叉树的头节点后面，再把2查到二叉树头节点后面……

构造树的时候要用到一个头节点，然而每次进入递归时使用的cur是同一个，导致后面的节点将前面的覆盖，我第一次写就出现了所有的左节点都消失了的现象

if (root->left)
{
    InOrder(root->left, cur);
}

我的解决办法是每次都找到树的最右节点再插入

while (cur->right) {cur = cur->right;}

最终代码如下

TreeNode* increasingBST(TreeNode* root) {
    if (root == NULL)
    {
        return root;
    }

    TreeNode* result = new TreeNode(0);
    TreeNode* current = result;
    InOrder(root, current);
    return current->right;
}

void InOrder(TreeNode* root, TreeNode* cur){
    if (root == NULL) return;

    if (root->left)
    {
        InOrder(root->left, cur);
    }

    TreeNode* node = new TreeNode(root->val);

    while (cur->right) {cur = cur->right;}
    cur->right = node;

    if (root->right)
    {
        InOrder(root->right, cur);
    }
}

LeetCode的讨论区还有一种先中序遍历，把结果存到vector里，再通过vector来生成树的做法，这样就简单很多，在此不提

下面是LeetCode目前最好的版本，分析一下

把class补全加上构造函数，涨知识了，还没在LeetCode上这么干过，这样就有了全局的树的头节点，递归的时候不用带上头节点了

class Solution
{
private:
    TreeNode *head, *track;

public:
    Solution()
    {
        head = new TreeNode(0);
        track = head;
    }

    TreeNode *increasingBST(TreeNode *root)
    {
        if(!root) return nullptr;
        increasingBST(root->left);
        track->right = new TreeNode(root->val);
        track = track->right;
        increasingBST(root->right);
        return head->right;
    }
};