Leetcode练习(Python):树类:第116题:填充每个节点的下一个右侧节点指针: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。
题目:
填充每个节点的下一个右侧节点指针: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。
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:
使用递归来解,分为左子树和右子树就好。主要是找规律。
程序:
""" # 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': if not root: return root if root and 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
程序2:层序
""" # 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': if not root: return root root.next = None current = root next_level_left = current.left while next_level_left: current.left.next = current.right if current.next: current.right.next = current.next.left current = current.next else: current.right.next = None current = next_level_left next_level_left = current.left return root