您的位置:首页 > Web前端 > Node.js

116. Populating Next Right Pointers in Each Node

2017-12-16 15:03 316 查看
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

/**
* 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) {
TreeLinkNode *p=root,*q;
if(root==nullptr)
return;
while(p->left /*&& p->right 因为是满二叉树,所以有left必有right*/){
q=p;
while(q){
q->left->next=q->right;
if(q->next)
q->right->next=q->next->left;
q=q->next;
}
p=p->left;
}
return;
}
};

class Solution {
public:
/*用递归空间复杂度不是常数级了*/
void connect(TreeLinkNode *root) {
if(!root)
return;
if(root->left && root->right)
root->left->next=root->right;
if(root->next && root->right)
root->right->next=root->next->left;
connect(root->left);
connect(root->right);
return;
}
};
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: