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【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
连接时,按照下面的规则进行就可以了:
left->next=father->right
right->next=father->next->left;
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 if(root==NULL) return;14 15 if(root->left!=NULL) root->left->next=root->right;16 if(root->right!=NULL)17 {18 if(root->next!=NULL) root->right->next=root->next->left;19 }20 connect(root->left);21 connect(root->right);22 23 }24 };25
【leetcode】Populating Next Right Pointers in Each Node
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