[LeetCode] 116. 填充每个节点的下一个右侧节点指针

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

题目描述:

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。

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":{"$id":"5","left":{"$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:

与下一题117. 填充每个节点的下一个右侧节点指针 II都可以用BFS方法,如下:

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

但是不符合题意, 要使用常量级额外空间

所以我们用其他空间(O(1))的方法

思路一: 递归

思路二: 迭代

代码:

思路一:

def connect(self, root: 'Node') -> 'Node':
        if not root:
            return 
        if root.left:
            root.left.next = root.right
            if root.next:
                root.right.next = root.next.left
        self.connect(root.left)
        self.connect(root.right)
        return root

java

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

思路二:

class Solution:
    def connect(self, root: 'Node') -> 'Node':
        pre = root
        while pre:
            cur = pre
            while cur:
                if cur.left: cur.left.next = cur.right
                if cur.right and cur.next: cur.right.next = cur.next.left
                cur = cur.next
            pre = pre.left
        return root

java

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