LeetCode: Populating Next Right Pointer in Each Node
LeetCode: Populating Next Right Pointer in Each Node
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
地址:https://oj.leetcode.com/problems/populating-next-right-pointers-in-each-node/
算法:用递归好简单有木有。按上面的例子,先完成左右子树的连接,然后,根的next指向空,第二个节点指向第三个节点,第五个节点指向第六个节点。代码:
/**
* 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 ;
root->next = NULL;
connect(root->left);
connect(root->right);
TreeLinkNode *p = root->left;
TreeLinkNode *q = root->right;
while(p && q){
p->next = q;
p = p->right;
q = q->left;
}
}
};
第二题:
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
地址:https://oj.leetcode.com/problems/populating-next-right-pointers-in-each-node-ii/
算法:这道题与前面一题不同的是,这里的二叉树不再是完美二叉树,而是一颗普通的二叉树。由于是一颗普通的二叉树,所以沿着右指针走不一定会到达下一层的最后一个节点,因为这个右指针可能为空;同样,沿着左指针走也不一定会到达下一层的第一个节点,因为这个左指针可能为空。所以,我们需要一个找到当前节点下一层的第一个节点的函数,然后顺着next也可以到达该层的最后一个节点。利用这样的函数,我们就可以完成题目的要求。代码:
/**
* 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 ;
root->next = NULL;
connect(root->left);
connect(root->right);
TreeLinkNode *p = root->left;
TreeLinkNode *q = root->right;
while(p && q){
TreeLinkNode *r = nextLevelFirst(p);
while(p->next){
p = p->next;
}
p->next = q;
p = r;
q = nextLevelFirst(q);
}
}
TreeLinkNode * nextLevelFirst(TreeLinkNode *p){
while(p){
if(p->left){
return p->left;
}else if(p->right){
return p->right;
}
p = p->next;
}
return NULL;
}
};
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