【leetcode】116. Populating Next Right Pointers in Each Node【java】
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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
/** * Definition for binary tree with next pointer. * public class TreeLinkNode { * int val; * TreeLinkNode left, right, next; * TreeLinkNode(int x) { val = x; } * } *///方法1:递归/*public class Solution { public void connect(TreeLinkNode root) { if (root == null || root.left == null){ return; } connectNode(root.left, root.right); } public void connectNode(TreeLinkNode node1, TreeLinkNode node2) { node1.next = node2; if (node1.left != null){ connectNode(node1.right, node2.left); connectNode(node1.left, node1.right); connectNode(node2.left, node2.right); } }}*///方法二:非递归public class Solution { public void connect(TreeLinkNode root) { if (root == null || root.left == null) { return; } TreeLinkNode start = root; while (start != null) { TreeLinkNode cur = start; while (cur != null) { if (cur.left != null) { cur.left.next = cur.right; } if (cur.right != null && cur.next != null) { cur.right.next = cur.next.left; } cur = cur.next; } start = start.left; } }}
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