转载

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:

1. You may only use constant extra space.
2. Recursive approach is fine, implicit stack space does not count as extra space for this problem.

难度：medium

题目：给定二叉树，计算结点指向其相邻右结点的下一跳指针。如果没有相邻的右结点则next为NULL.

思路：递归

Runtime: 3 ms, faster than 18.40% of Java online submissions for Populating Next Right Pointers in Each Node II.

Memory Usage: 51.8 MB, less than 100.00% of Java online submissions for Populating Next Right Pointers in Each Node II.

/*
// 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) {
            ptr = buildNodeNext(ptr);
        }
        
        return root;
    }
    
    private Node buildNodeNext(Node head) {
        if (null == head) {
            return null;
        }
        Node nextLevelLeftMost = buildNodeNext(head.next);
        Node left = head.left;
        Node right = head.right;
        if (left != null && right != null) {
            left.next = right;
            right.next = nextLevelLeftMost;
            return left;
        }
        
        if (left != null && right == null) {
            left.next = nextLevelLeftMost;
            return left;
        } 
        
        if (left == null && right != null) {
            right.next = nextLevelLeftMost;
            return right;
        }
        
        return nextLevelLeftMost;
    }
}

原文 https://segmentfault.com/a/1190000018179727

正文到此结束

所属分类： Java 编程技术

本文标签： java example App 递归 UI id IO https constant struct ACE node http root build
版权声明： 本文为互联网转载文章，出处已在文章中说明(部分除外)。如果侵权，请联系本站长删除，谢谢。
本文海报： 生成海报一生成海报二