方法一:层序遍历
解题思路:
和第116题一模一样,其实,用队列queue更简单一些,不用将顺序倒来倒去。通过使用队列的长度信息queue.size(),可以只需要一个队列就能做到层序遍历。
/*
// 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->right)
s2.push(temp1->right);
if(temp1->left)
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;
}
};