[LeetCode] 117. 填充每个节点的下一个右侧节点指针 II

题目链接 : https://leetcode-cn.com/problems/populating-next-right-pointers-in-each-node-ii/

题目描述:

给定一个二叉树

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。

示例:

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

解释：给定二叉树如图 A 所示，你的函数应该填充它的每个 next 指针，以指向其下一个右侧节点，如图 B 所示。

提示：

你只能使用常量级额外空间。
使用递归解题也符合要求，本题中递归程序占用的栈空间不算做额外的空间复杂度。

思路:

和上一题一样,可以用BFS做

def connect(self, root: 'Node') -> 'Node':
        from collections import deque
        if not root: return root
        queue = deque()
        queue.appendleft(root)
        while queue:
            p = None
            n = len(queue)
            for _ in range(n):
                tmp = queue.pop()
                if p:
                    p.next = tmp
                    p = p.next
                else:
                    p = tmp
                if tmp.left:
                    queue.appendleft(tmp.left)
                if tmp.right:
                    queue.appendleft(tmp.right)
            p.next = None 
        return root

但是题目要求我们用常数空间,我们借用上一题的迭代方法!

用cur记层该层访问节点

用head记录下一层合成链表的开头.

用tail记录下一层合成链表的结尾.

直接看代码, 就能理解了!

代码:

"""
# Definition for a Node.
class Node:
    def __init__(self, val, left, right, next):
        self.val = val
        self.left = left
        self.right = right
        self.next = next
"""
class Solution:
    def connect(self, root: 'Node') -> 'Node':
        cur = root
        head = None
        tail = None
        while cur:
            while cur:
                if cur.left:
                    if not head:
                        head = cur.left
                        tail = cur.left
                    else:
                        tail.next = cur.left
                        tail = tail.next
                if cur.right:
                    if not head:
                        head = cur.right
                        tail = cur.right
                    else:
                        tail.next = cur.right
                        tail = tail.next
                cur = cur.next
            cur = head
            head = None
            tail = None
        return root

java

class Solution {
    public Node connect(Node root) {
        Node cur = root;
        Node head = null;
        Node tail = null;
        while (cur != null) {
            while (cur != null) {
                if (cur.left != null) {
                    if (head == null) {
                        head = cur.left;
                        tail = head;
                    } else {
                        tail.next = cur.left;
                        tail = tail.next;
                    }
                }
                if (cur.right != null) {
                    if (head == null) {
                        head = cur.right;
                        tail = head;
                    } else {
                        tail.next = cur.right;
                        tail = tail.next;
                    }
                }
                cur = cur.next;
            }
            cur = head;
            head = null;
            tail = null;
        }
        return root;
    }
}