Populating Next Right Pointers in Each Node 更好的方法待补充
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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
第一感觉是用层序遍历,虽然ac了,但是空间复杂度似乎不满足常数的要求。
/** * 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 == 0) return; queue<TreeLinkNode *> q; q.push(root); q.push(0); while(!q.empty()) { TreeLinkNode * cur = q.front(); q.pop(); if(cur) { if(cur -> left) q.push(cur -> left); if(cur -> right) q.push(cur -> right); if(!q.empty()) if(q.front() == 0) cur -> next = NULL; else cur -> next = q.front(); } else if(!q.empty()) q.push(0); } }};
待补充
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