116. Populating Next Right Pointers in Each Node

题目：

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

链接：  http://leetcode.com/problems/populating-next-right-pointers-in-each-node/

4/16/2017

BB电面准备

抄答案的，递归方法，然而并不是constant extra space

 1 /**
 2  * Definition for binary tree with next pointer.
 3  * public class TreeLinkNode {
 4  *     int val;
 5  *     TreeLinkNode left, right, next;
 6  *     TreeLinkNode(int x) { val = x; }
 7  * }
 8  */
 9 public class Solution {
10     public void connect(TreeLinkNode root) {
11         if (root == null) return;
12         if (root.right != null) {
13             root.left.next = root.right;
14             if (root.next != null) {
15                 root.right.next = root.next.left;
16             }
17         }
18         connect(root.left);
19         connect(root.right);
20         return;
21     }
22 }

iterative，注意最后一步中需要head.next = null。一旦缺少这一步，cur->next在判断到root在最后一层时依然保持着这一层的next，会直接跳过前两个判断直接到root == null的判断，无限循环。有了head.next = null，当root为叶节点时，当判断完这一层root = cur.next，变为0，跳出外层循环。

head是root下一层的头, cur在root下一层按照next顺序来遍历。

 1 public class Solution {
 2     public void connect(TreeLinkNode root) {
 3         if (root == null) return;
 4         TreeLinkNode cur = new TreeLinkNode(-1);
 5         TreeLinkNode head = cur;
 6         
 7         while (root != null) {
 8             if (root.left != null) {
 9                 cur.next = root.left;
10                 cur = cur.next;
11             }
12             if (root.right != null) {
13                 cur.next = root.right;
14                 cur = cur.next;
15             }
16             root = root.next;
17             if (root == null) {
18                 cur = head;
19                 root = head.next;
20                 head.next = null;
21             }
22         }
23         return;
24     }
25 }

其他解法：

https://discuss.leetcode.com/topic/2202/a-simple-accepted-solution

 1 void connect(TreeLinkNode *root) {
 2     if (root == NULL) return;
 3     TreeLinkNode *pre = root;
 4     TreeLinkNode *cur = NULL;
 5     while(pre->left) {
 6         cur = pre;
 7         while(cur) {
 8             cur->left->next = cur->right;
 9             if(cur->next) cur->right->next = cur->next->left;
10             cur = cur->next;
11         }
12         pre = pre->left;
13     }
14 }

更多讨论：

https://discuss.leetcode.com/category/124/populating-next-right-pointers-in-each-node

posted @ 2017-04-25 10:03  panini  阅读(96)  评论(0)  编辑  收藏  举报