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
/** * Definition for binary tree with next pointer. * struct TreeLinkNode { * int val; * TreeLinkNode *left, *right, *next; * TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {} * }; */ class Solution { public: void connect(TreeLinkNode *root) { if(root == NULL) return; queue<TreeLinkNode*> use; use.push(root); while(!use.empty()) { int s = use.size(); for(int i=0; i<s; i++) { TreeLinkNode* t = use.front(); use.pop(); if(i == s-1) t->next = NULL; else t->next = use.front(); if(t->left != NULL) use.push(t->left); if(t->right != NULL) use.push(t->right); } } } };
LeetCode--Populating Next Right Pointers in Each Node II
原文:http://blog.csdn.net/shaya118/article/details/42713113