117. Populating Next Right Pointers in Each Node II
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
.
Example:
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":{"$id":"5","left":null,"next":null,"right":{"$id":"6","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":{"$id":"5","left":null,"next":null,"right":null,"val":7},"right":null,"val":5},"right":null,"val":4},"next":{"$id":"6","left":null,"next":null,"right":{"$ref":"5"},"val":3},"right":{"$ref":"4"},"val":2},"next":null,"right":{"$ref":"6"},"val":1} 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.
Note:
- You may only use constant extra space.
- Recursive approach is fine, implicit stack space does not count as extra space for this problem.
class Solution { public Node connect(Node root) { if(root == null) return null; Queue<Node> q = new LinkedList(); q.offer(root); while(!q.isEmpty()){ Node pre = null; int s = q.size(); for(int i = 0; i < s; i++){ Node cur = q.poll(); if(i > 0){ pre.next = cur; } pre = cur; if(cur.left != null) q.offer(cur.left); if(cur.right != null) q.offer(cur.right); } } return root; } }
利用一个栈将下一层的节点保存。通过
pre
指针把栈里的元素一个一个接起来。可用于116题