「力扣」第 114 题：二叉树展开为链表（中等）

• 题目链接

Java 代码：

class TreeNode {
    
    
    int val;
    TreeNode left;
    TreeNode right;

    TreeNode(int x) {
    
    
        val = x;
    }
}


public class Solution {
    
    

    private TreeNode parent;

    public void flatten(TreeNode root) {
    
    
        if (root == null) {
    
    
            return;
        }

        flatten(root.right);
        flatten(root.left);

        root.right = parent;
        root.left = null;

        // 暂存父亲节点
        parent = root;
    }
}

Python 代码：

class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right


class Solution:

    def __init__(self):
        self.parent = None

    def flatten(self, root: TreeNode) -> None:
        """
        Do not return anything, modify root in-place instead.
        """

        if root is None:
            return

        self.flatten(root.right)
        self.flatten(root.left)

        root.right = self.parent
        root.left = None

        self.parent = root