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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
分析:此题是level-order traversal的变形,与level-order traversal不同的是该题要求constant space。一般的level-order traversal用一个容器保存上一层的节点是O(n)space complexity,所以这里不同直接照搬。分析此题,一个节点p的左孩子的next是节点p的右孩子,p的节点的右孩子的next是p的next节点的左孩子,由这个我们便可以通过只维护节点p来完成populate。时间复杂度为O(n),空间复杂度为O(1)。代码如下:
class Solution {public: void connect(TreeLinkNode *root) { if(root == NULL) return; TreeLinkNode *head = root;//point to head of every level while(head->left && head->right){ TreeLinkNode *cur = head;//iterate through each node in a level while(cur){ cur->left->next = cur->right; cur->right->next = cur->next?cur->next->left:NULL; cur = cur->next; } head = head->left;//update head to left child of previous head } }};
Leetcode:Populating Next Right Pointers in Each Node
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