Leetcode79: Populating Next Right Pointers in Each Node
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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
不用运用额外的空间,但是同样可以用广度优先法一层一层的遍历各节点。
/** * 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) {} * }; */class Solution {public: void connect(TreeLinkNode *root) { if(!root) return; TreeLinkNode* Node; while(root->left != NULL) { Node = root; while(Node!= NULL) { Node->left->next = Node->right; if(Node->next != NULL) Node->right->next = Node->next->left; Node = Node->next; } root = root->left; } }};
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- Leetcode79: Populating Next Right Pointers in Each Node
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- Populating Next Right Pointers in Each Node
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- Populating Next Right Pointers in Each Node
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