leetcode 116. 填充每个节点的下一个右侧节点指针
问题描述
给定一个完美二叉树，其所有叶子节点都在同一层，每个父节点都有两个子节点。二叉树定义如下：
struct Node {
int val;
Node *left;
Node *right;
Node *next;
}
填充它的每个 next 指针，让这个指针指向其下一个右侧节点。如果找不到下一个右侧节点，则将 next 指针设置为 NULL。
初始状态下，所有 next 指针都被设置为 NULL。
示例：
输入：{"$id":"1","left":{"$id":"2","left":{"$id":"3","left":null,"next":null,"right":null,"val":4},"next":null,"right":{"$id":"4","left":null,"next":null,"right":null,"val":5},"val":2},"next":null,"right":{"$id":"5","left":{"$id":"6","left":null,"next":null,"right":null,"val":6},"next":null,"right":{"$id":"7","left":null,"next":null,"right":null,"val":7},"val":3},"val":1}
输出：{"$id":"1","left":{"$id":"2","left":{"$id":"3","left":null,"next":{"$id":"4","left":null,"next":{"$id":"5","left":null,"next":{"$id":"6","left":null,"next":null,"right":null,"val":7},"right":null,"val":6},"right":null,"val":5},"right":null,"val":4},"next":{"$id":"7","left":{"$ref":"5"},"next":null,"right":{"$ref":"6"},"val":3},"right":{"$ref":"4"},"val":2},"next":null,"right":{"$ref":"7"},"val":1}
解释：给定二叉树如图 A 所示，你的函数应该填充它的每个 next 指针，以指向其下一个右侧节点，如图 B 所示。
提示：
你只能使用常量级额外空间。
使用递归解题也符合要求，本题中递归程序占用的栈空间不算做额外的空间复杂度。
代码1（递归）
/*
// Definition for a Node.
class Node {
public:
int val;
Node* left;
Node* right;
Node* next;
Node() : val(0), left(NULL), right(NULL), next(NULL) {}
Node(int _val) : val(_val), left(NULL), right(NULL), next(NULL) {}
Node(int _val, Node* _left, Node* _right, Node* _next)
: val(_val), left(_left), right(_right), next(_next) {}
};
*/
class Solution {
public:
Node* connect(Node* root) {
if(!root || !root->left)return root;
root->left->next = root->right;
if(root->next){
root->right->next = root->next->left;
}
connect(root->left);
connect(root->right);
return root;
}
};
结果
执行用时 :36 ms, 在所有 C++ 提交中击败了43.33%的用户
内存消耗 :16.7 MB, 在所有 C++ 提交中击败了100.00%的用户
或者
/*
// Definition for a Node.
class Node {
public:
int val;
Node* left;
Node* right;
Node* next;
Node() : val(0), left(NULL), right(NULL), next(NULL) {}
Node(int _val) : val(_val), left(NULL), right(NULL), next(NULL) {}
Node(int _val, Node* _left, Node* _right, Node* _next)
: val(_val), left(_left), right(_right), next(_next) {}
};
*/
class Solution {
public:
Node* connect(Node* root) {
if(!root)return NULL;
if(root->left)root->left->next = root->right;
if(root->right)root->right->next = root->next?root->next->left:NULL;
connect(root->left);
connect(root->right);
return root;
}
};
代码2（层次遍历）
/*
// Definition for a Node.
class Node {
public:
int val;
Node* left;
Node* right;
Node* next;
Node() : val(0), left(NULL), right(NULL), next(NULL) {}
Node(int _val) : val(_val), left(NULL), right(NULL), next(NULL) {}
Node(int _val, Node* _left, Node* _right, Node* _next)
: val(_val), left(_left), right(_right), next(_next) {}
};
*/
class Solution {
public:
Node* connect(Node* root) {
if(!root)return NULL;
queue<Node*> q;
Node* tmp = NULL;
q.push(root);
int i,size;
while(!q.empty())
{
size = q.size();
for(i = 0; i < size; ++i)
{
tmp = q.front();
q.pop();
if(i < size-1)
{
tmp->next = q.front();
}
if(tmp->left)q.push(tmp->left);
if(tmp->right)q.push(tmp->right);
}
}
return root;
}
};
结果
执行用时 :36 ms, 在所有 C++ 提交中击败了43.33%的用户
内存消耗 :17.6 MB, 在所有 C++ 提交中击败了100.00%的用户
代码3（循环）
用两个指针 start 和 cur，其中 start 标记每一层的起始节点，cur 用来遍历该层的节点,从而避免使用额外空间。
/*
// Definition for a Node.
class Node {
public:
int val;
Node* left;
Node* right;
Node* next;
Node() : val(0), left(NULL), right(NULL), next(NULL) {}
Node(int _val) : val(_val), left(NULL), right(NULL), next(NULL) {}
Node(int _val, Node* _left, Node* _right, Node* _next)
: val(_val), left(_left), right(_right), next(_next) {}
};
*/
class Solution {
public:
Node* connect(Node* root) {
if(!root)return NULL;
Node* start = root,*cur = NULL;
while(start->left)
{
cur = start;
//从最左边的start开始遍历当前层
while(cur)
{
if(cur->left)cur->left->next = cur->right;
if(cur->next)cur->right->next = cur->next->left;
cur = cur->next;
}
start = start->left;
}
return root;
}
};
结果
执行用时 :28 ms, 在所有 C++ 提交中击败了79.48%的用户
内存消耗 :17.1 MB, 在所有 C++ 提交中击败了100.00%的用户
