1. 题目

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

2. 解题

2.1 递归

class Solution {
public:Node* connect(Node* root) {if(root == NULL || root->left == NULL)return root;root->left->next = root->right;if(root->next != NULL)root->right->next = root->next->left;connect(root->left);connect(root->right);return root;}
};

2.2 循环

class Solution {
public:Node* connect(Node* root) {if(root == NULL || root->left == NULL)return root;queue<Node*> q;q.push(root);Node *p;int n;while(!q.empty()){n = q.size();while(n--){p = q.front();if(p->left != NULL){q.push(p->left);q.push(p->right);p->left->next = p->right;if(p->next != NULL)p->right->next = p->next->left;}q.pop();}	}return root;}
};

2.3 O(1)空间复杂度

• 利用next指针

class Solution {
public:Node* connect(Node* root) {if(root == NULL || root->left == NULL)return root;Node *parent = root, *lvfirst, *prev;while(parent != NULL){lvfirst = parent->left;if(lvfirst != NULL){while(parent){parent->left->next = parent->right;prev = parent->right;parent = parent->next;if(parent)prev->next = parent->left;}}parent = lvfirst;	}return root;}
};