Leetcode Populating Next Right Pointers in Each Node
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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
public static void connect(TreeLinkNode root) { // 空节点就直接返回 if(root == null){ return; } // 左节点非空,连接右节点 if(root.left != null){ root.left.next = root.right; } // 借助root.next处理5连6的情况 if(root.right!=null && root.next!=null){ root.right.next = root.next.left; } connect(root.left); connect(root.right); }
public class Solution { 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); } }
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