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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:
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
Code:
1 void connect(TreeLinkNode *root) { 2 if (!root) 3 return ; 4 deque<TreeLinkNode*>a; 5 deque<TreeLinkNode*>b; 6 a.push_back(root); 7 8 while (!a.empty()) 9 { 10 while (!a.empty()) 11 { 12 TreeLinkNode* p = a.front(); 13 a.pop_front(); 14 if(p->left) 15 b.push_back(p->left); 16 if(p->right) 17 b.push_back(p->right); 18 19 if (a.empty()) 20 p->next = NULL; 21 else 22 p->next = a.front(); 23 } 24 a = b; 25 b.clear(); 26 } 27 }
Populating Next Right Pointers in Each Node II
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原文地址：http://www.cnblogs.com/happygirl-zjj/p/4590649.html
