You are given a perfect binary tree where all leaves are on the same level, and every parent has two children. The binary tree has the following definition:node
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.
Example:app
Input: {"$id":"1","left":{"$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":{"$id":"7","left":null,"next":null,"right":null,"val":7},"val":3},"val":1} Output: {"$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} Explanation: Given the above perfect binary tree (Figure A), your function should populate each next pointer to point to its next right node, just like in Figure B.
Note:
You may only use constant extra space.
Recursive approach is fine, implicit stack space does not count as extra space for this problem.less
難度:mediumthis
題目:給定一滿二叉樹,全部葉結點在同一層,每一個結點都有左右子樹。spa
思路:自上至下自左至右遍歷code
Runtime: 0 ms, faster than 100.00% of Java online submissions for Populating Next Right Pointers in Each Node.
Memory Usage: 36.1 MB, less than 100.00% of Java online submissions for Populating Next Right Pointers in Each Node.ci
/* // Definition for a Node. class Node { public int val; public Node left; public Node right; public Node next; public Node() {} 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 ptr = root; while (ptr != null) { Node head = ptr, rightMost = null; ptr = ptr.left; while (head != null && head.left != null) { if (rightMost != null) { rightMost.next = head.left; } head.left.next = head.right; rightMost = head.right; head = head.next; } } return root; } }