Given the following perfect binary tree,
1
/ 2 3
/ \ / 4 5 6 7
After calling your function, the tree should look like:
1 -> NULL
/ 2 -> 3 -> NULL
/ \ / 4->5->6->7 -> NULL
ver0：
1 class Solution { 2 public: 3 void connect(TreeLinkNode *root) { 4 if(root == NULL) return; 5 root->left->next = root->right;//ERROR 6 if(root->next) root->right->next = root->next->left; 7 connect(root->left); connect(root->right); 8 root->next = NULL; 9 } 10 };
RA:{0} root->left right均为NULL。
ver1：28ms
1 class Solution { 2 public: 3 void connect(TreeLinkNode *root) { 4 if(root == NULL) return; 5 if(root->left && root->right){ 6 root->left->next = root->right; 7 if(root->next) root->right->next = root->next->left; 8 connect(root->left); connect(root->right); 9 } 10 else return; 11 } 12 };
ver2:24ms
1 class Solution { 2 public: 3 void connect(TreeLinkNode *root) { 4 if(root == NULL) return; 5 if(root->left && root->right){ 6 root->left->next = root->right; 7 if(root->next) root->right->next = root->next->left; 8 9 } 10 connect(root->left); connect(root->right); 11 // else return; 12 } 13 };
LeetCode 116. Populating Next Right Pointers in Each Node
原文：http://www.cnblogs.com/co0oder/p/5196937.html