Populating Next Right Pointers in Each Node
Total Accepted: 72323 Total Submissions: 199207 Difficulty: Medium
Given a binary tree
struct TreeLinkNode {
TreeLinkNode *left;
TreeLinkNode *right;
TreeLinkNode *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.
Note:
- You may only use constant extra space.
- You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).
For example,
Given the following perfect binary tree,
1
/ \
2 3
/ \ / \
4 5 6 7
After calling your function, the tree should look like:
1 -> NULL
/ \
2 -> 3 -> NULL
/ \ / \
4->5->6->7 -> NULL
深度优先遍历，这题跟判断是否是镜像二叉树的思路是一样的。
/**
* 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 dfs(TreeLinkNode* root1,TreeLinkNode* root2){
if(!root1) return; root1->next = root2; dfs(root1->left,root1->right); if(!root2) return; dfs(root1->right,root2->left);
dfs(root2->left,root2->right);
}
void connect(TreeLinkNode *root) {
if(!root) return ;
dfs(root->left,root->right);
}
};
Next challenges: (H) Populating Next Right Pointers in Each Node II
Populating Next Right Pointers in Each Node II
Total Accepted: 51329 Total Submissions: 158800 Difficulty: Hard
Follow up for problem "Populating Next Right Pointers in Each Node".
What if the given tree could be any binary tree? Would your previous solution still work?
Note:
- You may only use constant extra space.
For example,
Given the following binary tree,
1
/ \
2 3
/ \ \
4 5 7
After calling your function, the tree should look like:
1 -> NULL
/ \
2 -> 3 -> NULL
/ \ \
4-> 5 -> 7 -> NULL 广度优先遍历
/**
* 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* head = NULL;
TreeLinkNode* tail = NULL;
TreeLinkNode* cur = root; while(cur){
if(cur->left){
if(tail){
tail->next = cur->left;
tail = cur->left;
}else{
tail = cur->left;
head = cur->left;
}
}
if(cur->right){
if(tail){
tail->next = cur->right;
tail = cur->right;
}else{
tail = cur->right;
head = cur->right;
}
}
cur = cur->next;
if(!cur){
cur = head;
tail = NULL;
head = NULL;
}
}
}
};