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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
/** * Definition for binary tree with next pointer. * struct TreeLinkNode { * int val; * TreeLinkNode *left, *right, *next; * TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {} * }; */ struct TreeLinkNode { int val; TreeLinkNode *left,*right,*next; TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {} }; class Solution { public: void connect(TreeLinkNode *root) { if(root == NULL) return ; if(root->left != NULL) root->left->next = root->right; if(root->right != NULL && root->next != NULL) root->right->next = root->next->left; connect(root->left); connect(root->right); } };
leetcode - Populating Next Right Pointers in Each Node
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