Leetcode 117. Populating Next Right Pointers in Each Node II

文章作者:Tyan
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1. Description

Populating Next Right Pointers in Each Node II

2. Solution

  • Version 1
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/**
* Definition for binary tree with next pointer.
* struct TreeLinkNode {
* int val;
* TreeLinkNode *left, *right, *next;
* TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
* };
*/
class Solution {
public:
void connect(TreeLinkNode *root) {
if(!root) {
return;
}
queue<TreeLinkNode*> q1;
q1.push(root);
queue<TreeLinkNode*> q2;
TreeLinkNode* pre = nullptr;
TreeLinkNode* current = nullptr;
while(!q1.empty()) {
current = q1.front();
q1.pop();
if(current->left) {
q2.push(current->left);
}
if(current->right) {
q2.push(current->right);
}
if(pre) {
pre->next = current;
}
pre = current;
if(q1.empty()) {
q1 = q2;
q2 = {};
pre = nullptr;
}
}
}
};
  • Version 2
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/**
* Definition for binary tree with next pointer.
* struct TreeLinkNode {
* int val;
* TreeLinkNode *left, *right, *next;
* TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
* };
*/
class Solution {
public:
void connect(TreeLinkNode *root) {
if(!root) {
return;
}
queue<TreeLinkNode*> q;
q.push(root);
connect(q);
}

private:
void connect(queue<TreeLinkNode*>& q) {
TreeLinkNode* pre = nullptr;
TreeLinkNode* current = nullptr;
queue<TreeLinkNode*> q2;
while(!q.empty()) {
current = q.front();
q.pop();
if(current->left) {
q2.push(current->left);
}
if(current->right) {
q2.push(current->right);
}
if(pre) {
pre->next = current;
}
pre = current;
}
if(!q2.empty()) {
connect(q2);
}
}
};
  • Version 3
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/**
* Definition for binary tree with next pointer.
* struct TreeLinkNode {
* int val;
* TreeLinkNode *left, *right, *next;
* TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
* };
*/
class Solution {
public:
void connect(TreeLinkNode *root) {
if(!root) {
return;
}
TreeLinkNode* parent = root;
TreeLinkNode* head = new TreeLinkNode(0);
while(root) {
parent = root;
TreeLinkNode* current = head;
while(parent) {
if(parent->left && parent->right) {
current->next = parent->left;
current->next->next = parent->right;
current = parent->right;
}
else if(parent->left) {
current->next = parent->left;
current = parent->left;
}
else if(parent->right){
current->next = parent->right;
current = parent->right;
}
parent = parent->next;
}
current->next = nullptr;
root = head->next;
}
delete head;
}
};

Reference

  1. https://leetcode.com/problems/populating-next-right-pointers-in-each-node-ii/description/
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