Populating Next Right Pointers in Each Node

Populating Next Right Pointers in Each Node

Total Accepted: 2529 Total
Submissions: 7506My Submissions

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

思路：层次遍历。同一层的节点连接成链表。用一个vector模拟队列，cur指向当前层的开始，last指向队列尾。

/**
* 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) {
// IMPORTANT: Please reset any member data you declared, as
// the same Solution instance will be reused for each test case.
if(root == NULL)
return;
vector < TreeLinkNode* > vec;
vec.push_back(root);
int cur = 0;
int last = vec.size();
while(cur < vec.size()){
last = vec.size();   //新的一行访问开始，重新定位last到当前行最后一个极点的下一个位置

while(cur < last ){
vec[cur]->next = cur < last-1 ? vec[cur+1] : NULL;   //当前节点指向同一层的下一个节点

if(vec[cur]->left != NULL)
vec.push_back(vec[cur]->left);
if(vec[cur]->right != NULL)
vec.push_back(vec[cur]->right);
cur++;
}
}
}
};

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