114.二叉树展开为链表

Posted 2020-12-11 cong12586

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.
        """
        return self.dfs(root,None)
    def dfs(self,root,Node):
        if not root : return Node
        # 记录当前遍历节点的前一个节点
        # 首先将右子树转换完成，然后将转换后的链表与左子树最后一层的
        # 最右边的节点
        Node = self.dfs(root.right,Node)
        Node = self.dfs(root.left,Node)
        # 拼接
        root.right = Node
        root.left = None
        return root