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LeetCode——Populating Next Right Pointers in Each Node
Given a binary tree
struct TreeLinkNode { TreeLinkNode *left; TreeLinkNode *right; TreeLinkNode *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
.
Note:
- You may only use constant extra space.
- You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).
For example,
Given the following perfect binary tree,
1 / 2 3 / \ / 4 5 6 7
After calling your function, the tree should look like:
1 -> NULL / 2 -> 3 -> NULL / \ / 4->5->6->7 -> NULL原题链接: https://oj.leetcode.com/problems/populating-next-right-pointers-in-each-node/
题目: 给定一个二叉树(假设是完全二叉树),把每个节点的next指针指向其右侧节点。
思路:首先想到的是,层序遍历树,在遍历的同时添加节点对右侧节点的指针。
另一种简洁的方法是采用递归来实现,间单直观。
public void connect(TreeLinkNode root) { if (root == null) return; if (root.left != null) root.left.next = root.right; if (root.right != null) root.right.next = root.next == null ? null : root.next.left; connect(root.left); connect(root.right); } // Definition for binary tree with next pointer. public class TreeLinkNode { int val; TreeLinkNode left, right, next; TreeLinkNode(int x) { val = x; } }
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