Populating Next Right Pointers in Each Node II
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
Runtime: 40ms.
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 if(!root) return; 13 14 TreeLinkNode pre(-1); //previous node in the level 15 for(TreeLinkNode* current = root, *move = ⪯ current; current = current->next){ 16 if(current->left){ 17 move->next = current->left; 18 move = move->next; 19 } 20 if(current->right){ 21 move->next = current->right; 22 move = move->next; 23 } 24 //current = current->next; //children of the current node have been dealt, move to the next node in the level 25 } 26 root = pre.next; //assign the first node in the next level to root and begin the new recursion 27 connect(root); 28 } 29 };
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