Problem description:
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 7After calling your function, the tree should look like:
1 -> NULL / 2 -> 3 -> NULL / 4-> 5 -> 7 -> NULL
Accepted Code:
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_helper(TreeLinkNode *root) {
12 if (root == NULL) return ;
13 // when root is a leaf node
14 if (root->left == NULL && root->right == NULL) return;
15 if (root->left != NULL && root->right != NULL)
16 root->left->next = root->right;
17
18 TreeLinkNode *p = NULL, *head = root;
19 // iterator on the level
20 while (head->next != NULL && head->next->left == NULL && head->next->right == NULL)
21 head = head->next;
22
23 if (head->next != NULL) {
24 if (head->next->left != NULL)
25 p = head->next->left;
26 else
27 p = head->next->right;
28 }
29 // right child is not null
30 if (root->right != NULL)
31 root->right->next = p;
32 else
33 root->left->next = p;
34 // recursive solve right and left child of root
35 // attention that the right child should be proceeded first
36 connect_helper(root->right);
37 connect_helper(root->left);
38 }
39 void connect(TreeLinkNode *root) {
40 // root = NULL
41 if (root == NULL) return ;
42 // deal root node
43 root->next = NULL;
44
45 connect_helper(root);
46 }
47 };