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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:
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
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 };
Populating Next Right Pointers in Each Node II
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原文地址:http://www.cnblogs.com/amazingzoe/p/4683662.html