[Java]LeetCode116. 填充同一层的兄弟节点 | 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.
• Recursive approach is fine, implicit stack space does not count as extra space for this problem.
• You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).

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

---

给定一个二叉树

struct TreeLinkNode {
  TreeLinkNode *left;
  TreeLinkNode *right;
  TreeLinkNode *next;
}

填充它的每一个 next 指针，让这个指针指向其下一个右侧节点。若是找不到下一个右侧节点，则将 next 指针设置为 NULL。

初始状态下，全部 next 指针都被设置为 NULL。

说明:

• 你只能使用额外常数空间。
• 使用递归解题也符合要求，本题中递归程序占用的栈空间不算作额外的空间复杂度。

示例:

给定二叉树，

     1
   /  \
  2    3
 / \    \
4   5    7

调用你的函数后，该二叉树变为：

     1 -> NULL
   /  \
  2 -> 3 -> NULL
 / \    \
4-> 5 -> 7 -> NULL

---

0ms

 1 /*
 2 // Definition for a Node.
 3 class Node {
 4     public int val;
 5     public Node left;
 6     public Node right;
 7     public Node next;
 8 
 9     public Node() {}
10 
11     public Node(int _val,Node _left,Node _right,Node _next) {
12         val = _val;
13         left = _left;
14         right = _right;
15         next = _next;
16     }
17 };
18 */
19 class Solution {
20     public Node connect(Node root) {
21         Node levelStart = root;
22         while (levelStart != null){
23             Node cur = levelStart;
24             while (cur != null){
25                 if (cur.left != null){
26                     cur.left.next = cur.right;
27                 }
28                 if (cur.right != null && cur.next != null){
29                     cur.right.next = cur.next.left;
30                 }
31                 cur = cur.next;
32             }
33             levelStart = levelStart.left;
34         }
35         return root;        
36     }
37 }

---

1ms

 1 class Solution {
 2     public Node connect(Node root) {
 3 
 4         if (root == null)
 5             return root;
 6 
 7         Queue<Node> queue = new LinkedList<>();
 8         queue.offer(root);
 9 
10         while (!queue.isEmpty()) {
11 
12             List<Integer> currentLayer = new ArrayList<>();
13             int layerSize = queue.size();
14             for (int i = 0; i < layerSize; i++) {
15 
16                 Node currentNode = queue.remove();
17                 currentLayer.add(currentNode.val);
18 
19                 if (currentNode.left != null)
20                     queue.add(currentNode.left);
21                 if (currentNode.right != null)
22                     queue.add(currentNode.right);
23                 
24                 if( i < layerSize-1 ) {
25                     currentNode.next = queue.peek();
26                 } else {
27                     currentNode.next = null;
28                 }
29             }
30         }
31         return root;
32     }
33 }

---

32380 kb 

 1 class Solution {
 2     public Node connect(Node root) {
 3         if (root == null) return null;
 4         
 5         Queue<Node> queue = new ArrayDeque<>();
 6         
 7         Node current;
 8         int remainingNodesAtCurrentLevel = 1;
 9         int nbNodesPreviousLevel = 1;
10         queue.offer(root);
11         while (!queue.isEmpty()) {
12             current = queue.poll();
13             remainingNodesAtCurrentLevel--;
14             
15             if (current.left != null) { //because it's perfect
16                 queue.offer(current.left);
17                 queue.offer(current.right);
18             }
19             
20             if (remainingNodesAtCurrentLevel == 0) {
21                 current.next = null;
22                 nbNodesPreviousLevel = nbNodesPreviousLevel * 2;
23                 remainingNodesAtCurrentLevel = nbNodesPreviousLevel;
24                 
25                 continue;
26             }
27             
28             System.out.println(current.val + " -- " + queue.peek().val);
29             
30             current.next = queue.peek();
31         }
32         
33         return root;
34     }
35 }