标签：point pop figure inpu val sum next turn pac
Given a binary tree
struct Node {
int val;
Node *left;
Node *right;
Node *next;
}
Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL.
Initially, all next pointers are set to NULL.
Follow up:
Example 1:
Input: root = [1,2,3,4,5,null,7] Output: [1,#,2,3,#,4,5,7,#] Explanation: Given the above binary tree (Figure A), your function should populate each next pointer to point to its next right node, just like in Figure B. The serialized output is in level order as connected by the next pointers, with ‘#‘ signifying the end of each level.
/* // Definition for a Node. class Node { public int val; public Node left; public Node right; public Node next; public Node() {} public Node(int _val) { val = _val; } public Node(int _val, Node _left, Node _right, Node _next) { val = _val; left = _left; right = _right; next = _next; } }; */ class Solution { public Node connect(Node root) { Node cur = root; Node prev = null; Node head = null; while (cur != null) { while (cur != null) { if (cur.left != null) { if (prev != null) { prev.next = cur.left; } else { head = cur.left; } prev = cur.left; } if (cur.right != null) { if (prev != null) { prev.next = cur.right; } else { head = cur.right; } prev = cur.right; } cur = cur.next; } cur = head; head = null; prev = null; } return root; } }
[LC] 117. Populating Next Right Pointers in Each Node II
标签：point pop figure inpu val sum next turn pac
原文地址：https://www.cnblogs.com/xuanlu/p/12185413.html
