问题描述:
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.
- Recursive approach is fine, implicit stack space does not count as extra space for this problem.
- You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).
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
解题思路:
这道题当然首先考虑层序遍历。
重要的一点是找到每一层的开头。
所以我采用的策略是用一个指针next指向每一层的开头,初始化时为root
当next与当前的节点为同一节点时,我们可以知道我们开始了新的一层,所以需要找到下一层的最左边的节点。
if(cur == next || next == NULL){
next = cur->left ? cur->left : cur->right;
}
需要注意的是,这一层的最左边的节点可能不是下一层最左边节点的父节点。
代码:
/** * 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; queue<TreeLinkNode*> q; q.push(root); TreeLinkNode* next = root; TreeLinkNode* prev = NULL; while(!q.empty()){ TreeLinkNode* cur = q.front(); q.pop(); if(prev) prev->next = cur; if(cur == next || next == NULL){ next = cur->left ? cur->left : cur->right; } if(cur->left) q.push(cur->left); if(cur->right) q.push(cur->right); prev = cur; if(!q.empty() && q.front() == next) prev = NULL; } } };