[LeetCode]Populating Next Right Pointers in Each Node
Populating Next Right Pointers in Each Node
Given a binary tree
struct TreeLinkNode { TreeLinkNode *left; TreeLinkNode *right; TreeLinkNode *next; }
Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL
.
Initially, all next pointers are set to NULL
.
Note:
- You may only use constant extra space.
- You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).
For example,
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
递归实现,注意到上一层在下一层之前完成next的指向。
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 || !root->left) return; 13 root->left->next = root->right; 14 if(root->next) 15 { 16 root->right->next = root->next->left; 17 } 18 connect(root->left); 19 connect(root->right); 20 } 21 };
还有一种方法就是,每次一次性遍历一整行。这样只要tree的最左边节点就可以了。
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 || !root->left) return; 13 TreeLinkNode* p = root; 14 while(p) 15 { 16 p->left->next = p->right; 17 if(p->next) 18 { 19 p->right->next = p->next->left; 20 } 21 p = p->next; 22 } 23 connect(root->left); 24 } 25 };