LeetCode 897 - Increasing Order Search Tree

https://leetcode.com/articles/increasing-order-search-tree/

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.

https://zxi.mytechroad.com/blog/tree/leetcode-897-increasing-order-search-tree/

修改指向的方式其实比较简单，使用prev指针一直指向了构造出来的这个新树的最右下边的节点，在中序遍历过程中把当前节点的左指针给设置为None，然后把当前节点放到新树的右下角，这样类似于一个越来越长的链表的构建过程。

  private TreeNode prev;

  

  public TreeNode increasingBST(TreeNode root) {

    TreeNode dummy = new TreeNode(0);

    this.prev = dummy;

    inorder(root);

    return dummy.right;

  }

  

  private void inorder(TreeNode root) {

    if (root == null) return;

    inorder(root.left);

    this.prev.right = root;

    this.prev = root;

    this.prev.left = null;

    inorder(root.right);

  }

We can perform the same in-order traversal as in Approach 1. During the traversal, we'll construct the answer on the fly, reusing the nodes of the given tree by cutting their left child and adjoining them to the answer.

Space Complexity: $O(H)$ in additional space complexity, where $H$ is the height of the given tree, and the size of the implicit call stack in our in-order traversal. 

  TreeNode cur;

  public TreeNode increasingBST(TreeNode root) {

    TreeNode ans = new TreeNode(0);

    cur = ans;

    inorder(root);

    return ans.right;

  }

  public void inorder(TreeNode node) {

    if (node == null)

      return;

    inorder(node.left);

    node.left = null;

    cur.right = node;

    cur = node;

    inorder(node.right);

  }

https://leetcode.com/problems/increasing-order-search-tree/discuss/165885/C%2B%2BJavaPython-Self-Explained-5-line-O(N)

    def increasingBST(self, root, tail = None):
        
        # if this null node was a left child, tail is its parent
        # if this null node was a right child, tail is its parent's parent
        if not root: return tail

        # recursive call, traversing left while passing in the current node as tail
        res = self.increasingBST(root.left, root)

        # we don't want the current node to have a left child, only a single right child
        root.left = None

        # we set the current node's right child to be tail
        # what is tail? this part is important
        # if the current node is a left child, tail will be its parent
        # else if the current node is a right child, tail will be its parent's parent
        root.right = self.increasingBST(root.right, tail)

        # throughout the whole algorithm, res is the leaf of the leftmost path in the original tree
        # its the smallest node and thus will be the root of the modified tree
        return res

The most important thing to notice is that when we traverse left, the root is passed in, but when we traverse right, the tail is passed in.
In other words, traversing left passes in the current node to the next iteration, while traversing right passes in the current node's parent.

I didn't use this condition of BST, and just inorder output the whole tree.

Straigh forward idea:
res = inorder(root.left) + root + inorder(root.right)

Several tips here:

1. I pass a tail part to the function, so it can link it to the last node.
   This operation takes O(1), instead of O(N).
   Otherwise the whole time complexity will be O(N^2).
2. Also, remember to set root.left = null.
   Otherwise it will be TLE for Leetcode to traverse your tree.
3. Should arrange the old tree, not create a new tree.
   The judgement won't take it as wrong answer, but it is.

    public TreeNode increasingBST(TreeNode root) {
        return increasingBST(root, null);
    }

    public TreeNode increasingBST(TreeNode root, TreeNode tail) {
        if (root == null) return tail;
        TreeNode res = increasingBST(root.left, root);
        root.left = null;
        root.right = increasingBST(root.right, tail);
        return res;
    }

public TreeNode increasingBST(TreeNode root) {
        Stack<TreeNode> stack = new Stack<>();
        TreeNode dummy = new TreeNode(0), p = dummy; 
        while (root != null || !stack.empty()) {
            if (root != null) {
                stack.push(root);
                root = root.left;
            } else {
                TreeNode cur = stack.pop();
                root = cur.right;
                cur.left = null;
                p.right = cur;
                p = p.right;
            }
        }
        return dummy.right; 
    }

Once we have traversed all the nodes in increasing order, we can construct new nodes using those values to form the answer.

• Time Complexity: $O(N)$, where $N$ is the number of nodes in the given tree.
• Space Complexity: $O(N)$, the size of the answer. 

  public TreeNode increasingBST(TreeNode root) {

    List<Integer> vals = new ArrayList();

    inorder(root, vals);

    TreeNode ans = new TreeNode(0), cur = ans;

    for (int v : vals) {

      cur.right = new TreeNode(v);

      cur = cur.right;

    }

    return ans.right;

  }

  public void inorder(TreeNode node, List<Integer> vals) {

    if (node == null)

      return;

    inorder(node.left, vals);

    vals.add(node.val);

    inorder(node.right, vals);

  }

https://www.codetd.com/article/3154023

    public TreeNode increasingBST(TreeNode root) {
        List<Integer> order = new ArrayList<>();
        Stack<TreeNode> s = new Stack<>();
        while(root != null || !s.isEmpty()){
            while(root != null){
                s.push(root);
                root = root.left;
            }
            TreeNode node = s.pop();
            order.add(node.val);
            if(node.right != null){
                root = node.right;
            }
        }
        
        TreeNode newRoot = null;
        TreeNode p = null;
        Iterator<Integer> iter = order.iterator();
        while(iter.hasNext()){
            if(newRoot == null){
                newRoot = new TreeNode(iter.next());
                p = newRoot;
            }
            else{
                p.right = new TreeNode(iter.next());
                p = p.right;
            }
        }
        return newRoot;
    }