LeetCode 117 Populating Next Right Pointers in Each Node II
Given a binary tree
struct Node {
int val;
Node *left;
Node *right;
Node *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
.
Solution
按层将下一层的节点按照左右子节点的顺序 \(push\) 进队列,此时的顺序就是 \(next\) 的顺序,所以直接循环即可
点击查看代码
/*
// 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 {
private:
queue<Node*> q;
public:
Node* connect(Node* root) {
if(!root) return NULL;
q.push(root);
while(!q.empty()){
int n = q.size();
Node* prev =NULL;
Node* tmp;
while(n--){
tmp = q.front(); q.pop();
if(prev) prev->next = tmp;
prev = tmp;
if(prev->left)q.push(prev->left);
if(prev->right)q.push(prev->right);
}
tmp->next = NULL;
}
return root;
}
};