[C++]LeetCode116. 填充同一層的兄弟節點 | Populating Next Right Pointers in Each Node
阿新 • • 發佈:2018-11-13
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.
- Recursive approach is fine, implicit stack space does not count as extra space for this problem.
- You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).
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
給定一個二叉樹
struct TreeLinkNode { TreeLinkNode *left; TreeLinkNode *right; TreeLinkNode *next; }
填充它的每個 next 指標,讓這個指標指向其下一個右側節點。如果找不到下一個右側節點,則將 next 指標設定為 NULL
。
初始狀態下,所有 next 指標都被設定為 NULL
。
說明:
- 你只能使用額外常數空間。
- 使用遞迴解題也符合要求,本題中遞迴程式佔用的棧空間不算做額外的空間複雜度。
示例:
給定二叉樹,
1 / \ 2 3 / \ \ 4 5 7
呼叫你的函式後,該二叉樹變為:
1 -> NULL / \ 2 -> 3 -> NULL / \ \ 4-> 5 -> 7 -> NULL
8ms
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->next && root->right) 14 root->right->next = root->next->left; 15 if(root->right) 16 root->left->next = root->right; 17 connect(root->left); 18 connect(root->right); 19 } 20 };
12ms
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 { 13 vector<TreeLinkNode*> tmp; 14 queue<TreeLinkNode*> q; 15 if(!root){return ;}; 16 q.push(root); 17 q.push(NULL); 18 19 while(!q.empty()) 20 { 21 TreeLinkNode* node = q.front(); 22 q.pop(); 23 24 if(!node) 25 { 26 linker(tmp); 27 tmp.clear(); 28 if(q.size() > 0){q.push(NULL); }; 29 30 }else{ 31 32 tmp.push_back(node); 33 if(node -> left != NULL){q.push(node -> left); }; 34 if(node -> right != NULL){q.push(node -> right); }; 35 36 } 37 } 38 39 } 40 41 void linker(vector<TreeLinkNode*>& nodes) 42 { 43 for(int i = 0; i < nodes.size() - 1; ++i) 44 { 45 nodes[i] -> next = nodes[i + 1]; 46 } 47 48 nodes.back() -> next = NULL; 49 } 50 };
16ms
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) 13 return; 14 //佇列 15 queue<TreeLinkNode*> q; 16 q.push(root); 17 while (!q.empty()) 18 { 19 int len = q.size(); 20 cout << "len:" << len << endl; 21 for (int i = 0; i < len; i++) 22 { 23 TreeLinkNode* cur_node = q.front(); 24 q.pop(); 25 //當前結點的next,就是q在pop後的front結點 26 //注意:需要用len-1判斷,而不能用q.empty()判斷,因為q是在動態增長的 27 if (i < len - 1) 28 { 29 cur_node->next = q.front(); 30 } 31 //左右子樹入佇列 32 if (cur_node->left) 33 { 34 q.push(cur_node->left); 35 } 36 if (cur_node->right) 37 { 38 q.push(cur_node->right); 39 } 40 } 41 } 42 return; 43 } 44 };