# 114. Flatten Binary Tree to Linked List
# Medium

# 4387

# 412

# Add to List

# Share
# Given the root of a binary tree, flatten the tree into a "linked list":

# The "linked list" should use the same TreeNode class where the right child pointer points to the next node in the list and the left child pointer is always null.
# The "linked list" should be in the same order as a pre-order traversal of the binary tree.
 

# Example 1:


# Input: root = [1,2,5,3,4,null,6]
# Output: [1,null,2,null,3,null,4,null,5,null,6]
# Example 2:

# Input: root = []
# Output: []
# Example 3:

# Input: root = [0]
# Output: [0]
 

# Constraints:

# The number of nodes in the tree is in the range [0, 2000].
# -100 <= Node.val <= 100
 

# Follow up: Can you flatten the tree in-place (with O(1) extra space)?


# This solution works:

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def flatten(self, root: TreeNode) -> None:
        """
        Do not return anything, modify root in-place instead.
        """
        '''
        use stack !!!
        '''
        prev = None
        stack = [root]
        while stack:
            cur = stack.pop()
            if cur is None:
                continue
            if prev is not None:
                prev.right = cur
                prev.left = None
            stack.append(cur.right)
            stack.append(cur.left)
            prev = cur
        if prev is not None:
            prev.right = None
            prev.left = None
        return root
d487a50bd76de15ce2a4cae4becd557f8ef736f6
<mask token>

class Solution:
    <mask token>

class Solution:

    def flatten(self, root: TreeNode) ->None:
        """
        Do not return anything, modify root in-place instead.
        """
        """
        use stack !!!
        """
        prev = None
        stack = [root]
        while stack:
            cur = stack.pop()
            if cur is None:
                continue
            if prev is not None:
                prev.right = cur
                prev.left = None
            stack.append(cur.right)
            stack.append(cur.left)
            prev = cur
        if prev is not None:
            prev.right = None
            prev.left = None
        return root

# 114. Flatten Binary Tree to Linked List
# Medium

# 4387

# 412

# Add to List

# Share
# Given the root of a binary tree, flatten the tree into a "linked list":

# The "linked list" should use the same TreeNode class where the right child pointer points to the next node in the list and the left child pointer is always null.
# The "linked list" should be in the same order as a pre-order traversal of the binary tree.
 

# Example 1:


# Input: root = [1,2,5,3,4,null,6]
# Output: [1,null,2,null,3,null,4,null,5,null,6]
# Example 2:

# Input: root = []
# Output: []
# Example 3:

# Input: root = [0]
# Output: [0]
 

# Constraints:

# The number of nodes in the tree is in the range [0, 2000].
# -100 <= Node.val <= 100
 

# Follow up: Can you flatten the tree in-place (with O(1) extra space)?


# This solution works:

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def flatten(self, root: TreeNode) -> None:
        """
        Do not return anything, modify root in-place instead.
        """
        '''
        use stack !!!
        '''
        prev = None
        stack = [root]
        while stack:
            cur = stack.pop()
            if cur is None:
                continue
            if prev is not None:
                prev.right = cur
                prev.left = None
            stack.append(cur.right)
            stack.append(cur.left)
            prev = cur
        if prev is not None:
            prev.right = None
            prev.left = None
        return root