方法一:层序遍历(这是比较暴力愚蠢的方法)
解题思路:
这里使用层序遍历访问这颗完美二叉树,使用的是两个栈,而不是两个队列,因为这样在遍历每一层并指定next指针时会更方便一些,但是要注意的一点是,循环内的临时栈s2 是先push右子节点,再push左子节点。s2中节点顺序是反过来的,所以要依次pop到s1中。
/*
// Definition for a Node.
class Node {
public:
int val;
Node* left;
Node* right;
Node* next;
Node() {}
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 == NULL) return root;
//层序遍历
stack<Node*> s1;
s1.push(root);
while(!s1.empty()){
stack<Node*> s2;
Node* temp1 = NULL;
Node* temp2 = NULL;
while(!s1.empty()){
temp1 = s1.top();
s1.pop();
temp1->next = temp2;
temp2 = temp1;
if(temp1->left){
s2.push(temp1->right);
s2.push(temp1->left);
}
}
//s2中节点顺序是反过来的
while(!s2.empty()){
s1.push(s2.top());
s2.pop();
}
}
return root;
}
};
方法二:技巧型
解题思路:
因为是一颗完美二叉树,所以,对于某个节点来说,其左子节点的next指针直接指向其右子节点即可,右子节点指向其父节点next指针所指节点的左子节点,若其父节点next指针为null,则该右子节点的next也指向null。
但是递归耗时费内存啊!
/*
// Definition for a Node.
class Node {
public:
int val;
Node* left;
Node* right;
Node* next;
Node() {}
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 == NULL) return root;
if(root->left) root->left->next = root->right;
if(root->right) {
if(root->next) root->right->next = root->next->left;
}
root->left = connect(root->left);
root->right = connect(root->right);
return root;
}
};