[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
层次遍历
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) { 13 return; 14 } 15 queue<TreeLinkNode* > q; 16 TreeLinkNode *p; 17 int idx = 1, n; 18 q.push(root); 19 while (!q.empty()) { 20 n = idx - 1; 21 idx = 0; 22 p = q.front(); 23 if (q.front()->left != NULL) { 24 q.push(q.front()->left); 25 idx++; 26 } 27 if (q.front()->right != NULL) { 28 q.push(q.front()->right); 29 idx++; 30 } 31 q.pop(); 32 for (int i = 0; i < n; ++i) { 33 p->next = q.front(); 34 if (q.front()->left != NULL) { 35 q.push(q.front()->left); 36 idx++; 37 } 38 if (q.front()->right != NULL) { 39 q.push(q.front()->right); 40 idx++; 41 } 42 p = p->next; 43 q.pop(); 44 } 45 p->next = NULL; 46 } 47 } 48 };