为有牺牲多壮志,敢教日月换新天。

[Java]LeetCode117. 填充同一层的兄弟节点 II | Populating Next Right Pointers in Each Node II

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤微信公众号:山青咏芝(shanqingyongzhi)
➤博客园地址:山青咏芝(https://www.cnblogs.com/strengthen/
➤GitHub地址:https://github.com/strengthen/LeetCode
➤原文地址:https://www.cnblogs.com/strengthen/p/9953121.html 
➤如果链接不是山青咏芝的博客园地址,则可能是爬取作者的文章。
➤原文已修改更新!强烈建议点击原文地址阅读!支持作者!支持原创!
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

热烈欢迎,请直接点击!!!

进入博主App Store主页,下载使用各个作品!!!

注:博主将坚持每月上线一个新app!!!

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.

Example:

Given the following binary tree,

     1
   /  \
  2    3
 / \    \
4   5    7

After calling your function, the tree should look like:

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

给定一个二叉树

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

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 NULL

初始状态下,所有 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做额外的空间复杂度。

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 node = root;
22          while(node != null){
23             Node tempChild = new Node(0);
24             Node currentChild = tempChild;
25             while(node!=null){
26                 if(node.left != null) { currentChild.next = node.left; currentChild = currentChild.next;}
27                 if(node.right != null) { currentChild.next = node.right; currentChild = currentChild.next;}
28                 node = node.next;
29             }
30             node = tempChild.next;
31         }
32         return root;
33     }
34 }

1ms

 1 class Solution {
 2     public Node connect(Node root) {
 3         connect(root, null);
 4         return root;
 5     }
 6     
 7     private void connect(Node root, Node nxtRight) {
 8         if(root == null) {
 9             return;
10         }
11         
12         root.next = nxtRight;
13         Node nxtRightForRight = null;
14         Node curr = root.next;
15         while(curr != null) {
16             if(curr.left != null) {
17                 nxtRightForRight = curr.left;
18                 break;
19             } else if(curr.right != null) {
20                 nxtRightForRight = curr.right;
21                 break;
22             }
23             curr = curr.next;
24         }
25         connect(root.right, nxtRightForRight);
26         connect(root.left, root.right == null ? nxtRightForRight : root.right);
27         return;        
28     }
29 }

2ms

 1 class Solution {
 2     public Node connect(Node root) {
 3         if (root == null) {
 4             return root;
 5         }
 6         return connectHelper(root,0,new HashMap<Integer,Node>());//map put the last node of every level
 7     }
 8     
 9     private Node connectHelper(Node node,int level,Map<Integer,Node> map) {
10         
11         if (node == null) {
12             return node;
13         }
14         Node last = null;
15         if (map.containsKey(level)) {
16             last = map.get(level);            
17             last.next = node;
18         }
19         map.put(level,node);
20         Node left = connectHelper(node.left,level+1,map);
21         Node right = connectHelper(node.right,level+1,map); 
22         
23         return node;
24     }
25 }

3ms

 1 class Solution {
 2     public Node connect(Node root) {
 3         if (root == null) {
 4             return root;
 5         }
 6         Deque<Node> queue = new LinkedList<>();
 7         queue.offer(root);
 8         while (!queue.isEmpty()) {
 9             int size = queue.size();
10             Node prev = null,cur = null;
11             for (int i = 0; i < size; i++) {
12                 cur = queue.poll();
13                 if (prev != null) {
14                     prev.next = cur;
15                 }
16                 prev = cur;
17                 
18                 if (cur.left != null) {
19                     queue.offer(cur.left);
20                 }
21                 if (cur.right != null) {
22                     queue.offer(cur.right);
23                 }
24             }
25             cur.next = null;
26         }
27         return root;
28     } 
29 }

4ms

 1 class Solution {
 2     public Node connect(Node root) {
 3          if(root==null){
 4             return null;
 5         }
 6        LinkedList<Node> stack=new LinkedList<>();
 7        LinkedList<Node> temp=new LinkedList<>(); 
 8        List<List<Node>> list=new ArrayList<List<Node>>();
 9         List<Node> l=new ArrayList<>();
10         stack.addFirst(root);
11         while(!stack.isEmpty()){
12             Node rp=stack.removeFirst();
13             l.add(rp);
14             if(rp.left!=null){
15                 temp.addLast(rp.left);
16             }
17             if(rp.right!=null){
18                 temp.addLast(rp.right);
19             }
20             if(stack.isEmpty()){
21                 stack=temp;
22                 temp=new LinkedList<>();
23                 list.add(l);
24                 l=new ArrayList<>();
25             }
26         }
27         for(int i=0;i<list.size();i++){
28             for(int j=0;j<list.get(i).size();j++){
29                 if(j==list.get(i).size()-1){
30                     Node s=list.get(i).get(j);
31                     s.next=null;
32                     break;
33                 }
34                 Node a=list.get(i).get(j);
35                 a.next=list.get(i).get(j+1);
36             }
37         }
38         return root;
39     }
40 }

5ms

 1 class Solution {
 2     public Node connect(Node root) {
 3         if (root == null) {
 4             return null;
 5         }
 6         
 7         Node lastHead = root;
 8         Node curHead = lastHead.left != null ? lastHead.left : lastHead.right;
 9         
10         Node last = lastHead;
11         Node cur = curHead;
12         
13         while (cur != null) {
14             Node nxt = null;
15             while (last != null && nxt == null) {
16                 if (cur == last.left) {
17                     if (last.right != null) {
18                         nxt = last.right;
19                     } else {
20                         last = last.next;
21                     }
22                 } else if (cur == last.right) {
23                     last = last.next;
24                 } else {
25                     if (last.left != null) {
26                         nxt = last.left;
27                     } else if (last.right != null) {
28                         nxt = last.right;
29                     } else {
30                         last = last.next;
31                     }
32                 }
33             }
34             
35             cur.next = nxt;
36             cur = cur.next;
37             
38             // System.out.println(cur != null ? cur.val + " " : "null ");
39             if (cur == null) {
40                 lastHead = curHead;
41                 while (lastHead != null && lastHead.left == null && lastHead.right == null) {
42                     lastHead = lastHead.next;
43                 }
44                 
45                 // System.out.println("       " + lastHead.left + " " + lastHead.right + " " + lastHead.next.val);
46                 
47                 if (lastHead == null) {
48                     break;
49                 }
50                 curHead = lastHead.left != null ? lastHead.left : lastHead.right;
51                 
52                 last = lastHead;
53                 cur = curHead;
54                 if (cur != null) {
55                     System.out.println(lastHead.val + " " + curHead.val);    
56                 }                
57             }
58         }        
59         return root;        
60     }
61 }

 

posted @ 2018-11-13 16:54  为敢技术  阅读(260)  评论(0编辑  收藏  举报