LEETCODE: 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
感觉这个层层遍历的方式,用了N次,刷下一轮的时候,要想办法换个新的。
/** * 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 == NULL) return; queue<TreeLinkNode *> level; level.push(root); while(!level.empty()) { queue<TreeLinkNode *> newlevel; while(!level.empty()) { TreeLinkNode *front = level.front(); if(front->left != NULL) newlevel.push(front->left); if(front->right != NULL) newlevel.push(front->right); level.pop(); if(!level.empty()) front->next = level.front(); } level = newlevel; } }};
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