[LeetCode] 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
思路:前序的递归思路。时间复杂度O(n),,空间复杂度O(1)
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 == NULL) return; 13 if (root->left) { 14 root->left->next = root->right; 15 if (root->next) 16 root->right->next = root->next->left; 17 } 18 19 connect(root->left); 20 connect(root->right); 21 } 22 };
思路二:维护两个指针。
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 == NULL) return; 13 TreeLinkNode *pre = root; 14 TreeLinkNode *cur = NULL; 15 while (pre->left) { 16 cur = pre; 17 while (cur) { 18 cur->left->next = cur->right; 19 if (cur->next) cur->right->next = cur->next->left; 20 cur = cur->next; 21 } 22 23 pre = pre->left; 24 } 25 } 26 };