Populating Next Right Pointers in Each Node - LeetCode
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
Thoughts:
1.如果有左右节点,那么左节点的next指向右节点
2.如果有next,那么右节点的next指向next节点的左节点
3.递归遍历整棵树
/** * Definition for binary tree with next pointer. * public class TreeLinkNode { * int val; * TreeLinkNode left, right, next; * TreeLinkNode(int x) { val = x; } * } */ public class Solution { public void connect(TreeLinkNode root) { if(root==null) return; if(root.left!=null&&root.right!=null){ root.left.next=root.right; if(root.next!=null){ root.right.next=root.next.left; }
} connect(root.left); connect(root.right); } }