116. 填充每个节点的下一个右侧节点指针
题目描述:
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。
算法:递归解法
/*
// 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)
return NULL;
if(root->left)
root->left->next = root->right;
if(root->right)
root->right->next =root->next!=NULL? root->next->left : NULL;
connect(root->left);
connect(root->right);
return root;
}
};
算法2:
层序遍历的非递归解法,需要借助辅助队列
/*
// 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) return NULL;
queue<Node*> q;
q.push(root);
while (!q.empty()) {
int size = q.size();
for (int i = 0; i < size; ++i) {
Node *t = q.front(); q.pop();
if (t->left)
{
q.push(t->left);
t->left->next = t->right;
}
if (t->right)
{
q.push(t->right);
t->right->next = (t->next==NULL ? NULL : t->next->left);
}
}
}
return root;
}
};
另一种写法:
/*
// 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) return NULL;
queue<Node*> q;
q.push(root);
while (!q.empty()) {
int size = q.size();
for (int i = 0; i < size; ++i) {
Node *t = q.front(); q.pop();
if(i<size-1)
t->next = q.front();
if (t->left)
{
q.push(t->left);
}
if (t->right)
{
q.push(t->right);
}
}
}
return root;
}
};
算法3:
参考 博主题解
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