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

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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.

Example:

Given the following binary tree,

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，

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 }

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原文地址：https://www.cnblogs.com/strengthen/p/9953121.html