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114. Flatten Binary Tree to Linked List ​

题目 ​

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)?
Related Topics
  • 栈
  • 树
  • 深度优先搜索
  • 链表
  • 二叉树

  • 👍 1863
  • 👎 0

    • 思路 ​

    使用栈存储 right 节点，先入后出原理，依次添加到处理完 left 的子节点上

    python
    class Solution:
    def flatten(self, root: Optional[TreeNode]) -> None:
        """
        Do not return anything, modify root in-place instead.
        """
        if not root:
            return

        node = root
        stack = []

        while node:
            if node.right:
                stack.append(node.right)

            if node.left:
                node.right = node.left
                node.left = None
            elif stack:
                node.right = stack.pop()

            node = node.right
    py
    # leetcode submit region begin(Prohibit modification and deletion)
# 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: Optional[TreeNode]) -> None:
        """
        Do not return anything, modify root in-place instead.
        """
        if not root:
            return

        node = root

        while node:
            if node.left:
                predecessor = node.left
                while predecessor.right:
                    predecessor = predecessor.right

                predecessor.right = node.right
                node.right = node.left
                node.left = None

            node = node.right
















# leetcode submit region end(Prohibit modification and deletion)

    复杂度分析 ​

    • 时间复杂度 O(N)
    • 空间复杂度 O(N)

    空间复杂度为为1的解法 ​

    解法1的移动方式：

    自下到上的节点移动，故需要存储未移动的遍历节点

    解法2 的移动方式

    自上到下的节点移动，right 节点需要存储在 left 树的 rightmost 节点的 right 节点

    • 寻找 left 树的 rightmost 节点
    • 移动 right -> rightmost.right, 则 left -> right 节点
    • loop node.right 重复以上过程
    py
    # leetcode submit region begin(Prohibit modification and deletion)
# 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: Optional[TreeNode]) -> None:
        """
        Do not return anything, modify root in-place instead.
        """
        if not root:
            return

        node = root

        while node:
            if node.left:
                predecessor = node.left
                while predecessor.right:
                    predecessor = predecessor.right

                predecessor.right = node.right
                node.right = node.left
                node.left = None

            node = node.right
















# leetcode submit region end(Prohibit modification and deletion)

    复杂度分析 ​

    • 时间复杂度 O(N)
    • 空间复杂度 O(1)