116. 填充每个节点的下一个右侧节点指针

题目

介绍

给定一个完美二叉树，其所有叶子节点都在同一层，每个父节点都有两个子节点。二叉树定义如下：

struct Node {
    int val;
    Node *left;
    Node *right;
    Node *next;
}
填充它的每个 next 指针，让这个指针指向其下一个右侧节点。如果找不到下一个右侧节点，则将 next 指针设置为 NULL。

初始状态下，所有 next 指针都被设置为 NULL。

输入：{"$id":"1","left":{"$id":"2","left":{"$id":"3","left":null,"next":null,"right":null,"val":4},"next":null,"right":{"$id":"4","left":null,"next":null,"right":null,"val":5},"val":2},"next":null,"right":{"$id":"5","left":{"$id":"6","left":null,"next":null,"right":null,"val":6},"next":null,"right":{"$id":"7","left":null,"next":null,"right":null,"val":7},"val":3},"val":1}

输出：{"$id":"1","left":{"$id":"2","left":{"$id":"3","left":null,"next":{"$id":"4","left":null,"next":{"$id":"5","left":null,"next":{"$id":"6","left":null,"next":null,"right":null,"val":7},"right":null,"val":6},"right":null,"val":5},"right":null,"val":4},"next":{"$id":"7","left":{"$ref":"5"},"next":null,"right":{"$ref":"6"},"val":3},"right":{"$ref":"4"},"val":2},"next":null,"right":{"$ref":"7"},"val":1}

解法

方法一

思路

1. 每个 node 左子树的 next , 就是 node 的右子树
2. 每个 node 右子树的 next, 就是 node next 的 左子树

代码

    """
    # Definition for a Node.
    class Node:
        def __init__(self, val: int = 0, left: 'Node' = None, right: 'Node' = None, next: 'Node' = None):
            self.val = val
            self.left = left
            self.right = right
            self.next = next
    """

    class Solution:
        def connect(self, root: 'Node') -> 'Node':
            self.dfs(root,None)
            return root
        

        def dfs(self,root,p):
            if root:
                root.next = p
                self.dfs(root.left,root.right)
                if root.next:
                    self.dfs(root.right,root.next.left)
                else:
                    self.dfs(root.right,None)

复杂度计算

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