Populating Next Right Pointers in Each Node II

最新推荐文章于 2021-01-29 16:34:55 发布

weixin_30709061

最新推荐文章于 2021-01-29 16:34:55 发布

阅读量54

点赞数

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/zhengjiankang/p/3680276.html

本文提供了一种在二叉树中连接相邻节点的方法，并通过迭代的方式实现，仅使用常数额外空间。此外，还探讨了当输入为普通二叉树而非完美二叉树时解决方案的有效性。

 1 /**
 2  * Definition for binary tree with next pointer.
 3  * struct TreeLinkNode {
 4  *  int val;
 5  *  TreeLinkNode *left, *right, *next;
 6  *  TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
 7  * };
 8  */
 9 class Solution {
10 public:
11     void connect(TreeLinkNode *root) {
12         TreeLinkNode* cur = root;
13         while(cur) {
14             TreeLinkNode* node = cur;
15             TreeLinkNode* last = NULL;
16             cur = NULL;
17             while(node) {
18                 TreeLinkNode* left = node->left;
19                 TreeLinkNode* right = node->right;
20                 if(left || right) {
21                     if(last) last->next = left ? left : right;
22                     if(left) left->next = right;
23                     if(!cur) cur = left ? left : right;
24                     last = right ? right : left;
25                 }
26                 node = node->next;
27             }
28         }
29     }
30 };

Follow up for problem "Populating Next Right Pointers in Each Node".
What if the given tree could be any binary tree? Would your previous solution still work?
Note:
You may only use constant extra space.
For 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

Solution: 1. iterative way with CONSTANT extra space.
2. iterative way + queue. Contributed by SUN Mian(孙冕).
3. tail recursive solution.
*/

转载于:https://www.cnblogs.com/zhengjiankang/p/3680276.html