populating-next-right-pointers-in-each-node-ii

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Algorithms

Given a binary tree

<pre>
struct Node {
 int val;
 Node *left;
 Node *right;
 Node *next;
}
</pre>

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 <code>NULL</code>.

Initially, all next pointers are set to <code>NULL</code>.

&nbsp;
Example 1:
<img alt="" src="https://assets.leetcode.com/uploads/2019/02/15/117_sample.png" style="width: 500px; height: 171px;" />
<pre>
Input: root = [1,2,3,4,5,null,7]
Output: [1,#,2,3,#,4,5,7,#]
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. The serialized output is in level order as connected by the next pointers, with &#39;#&#39; signifying the end of each level.
</pre>

Example 2:

<pre>
Input: root = []
Output: []
</pre>

&nbsp;
Constraints:

<ul>
	<li>The number of nodes in the tree is in the range <code>[0, 6000]</code>.</li>
	<li><code>-100 &lt;= Node.val &lt;= 100</code></li>
</ul>

&nbsp;
Follow-up:

<ul>
	<li>You may only use constant extra space.</li>
	<li>The recursive approach is fine. You may assume implicit stack space does not count as extra space for this problem.</li>
</ul>

connect

Populating Next Right Pointers in Each Node II

Populating Next Right Pointers in Each Node

populating-next-right-pointers-in-each-node

填充每个节点的下一个右侧节点指针

Tree

Depth-First Search

Breadth-First Search

Linked List

Binary Tree

给定一个二叉树：

<pre>
struct Node {
 int val;
 Node *left;
 Node *right;
 Node *next;
}</pre>

填充它的每个 next 指针，让这个指针指向其下一个右侧节点。如果找不到下一个右侧节点，则将 next 指针设置为 <code>NULL</code> 。

初始状态下，所有&nbsp;next 指针都被设置为 <code>NULL</code> 。

&nbsp;

示例 1：
<img alt="" src="https://assets.leetcode.com/uploads/2019/02/15/117_sample.png" style="width: 500px; height: 171px;" />
<pre>
输入：root = [1,2,3,4,5,null,7]
输出：[1,#,2,3,#,4,5,7,#]
解释：给定二叉树如图 A 所示，你的函数应该填充它的每个 next 指针，以指向其下一个右侧节点，如图 B 所示。序列化输出按层序遍历顺序（由 next 指针连接），'#' 表示每层的末尾。</pre>

示例 2：

<pre>
输入：root = []
输出：[]
</pre>

&nbsp;

提示：

<ul>
	<li>树中的节点数在范围 <code>[0, 6000]</code> 内</li>
	<li><code>-100 &lt;= Node.val &lt;= 100</code></li>
</ul>

进阶：

<ul>
	<li>你只能使用常量级额外空间。</li>
	<li>使用递归解题也符合要求，本题中递归程序的隐式栈空间不计入额外空间复杂度。</li>
</ul>

<ul>
</ul>