題目鏈接:Populating Next Right Pointers in Each Node
題目內容:
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
題目解法:
這道題本來只是一道簡單的計層BFS題目,但是有一個對於常數額外空間複雜度的要求,因此不能使用傳統的隊列BFS,可以只定義一個變量cur,根據樹的性質來完成BFS,具體方法如下:
首先初始化cur爲root,cur的每次動作是從左到右的,通過cur = cur->next實現;當cur->next==NULL時,讓cur沿着root的左子樹往下走,也就是root = root->left; cur = root;當我們設置next時,實際上我們是設置的下一層的next,因此當訪問當前層的next時,已經被設置過。
下面舉個例子,如上面一棵完全二叉樹,假設現在cur=2,在第一層我們已經設置了2的next爲3,因此在本層通過cur->left->next=cur->right設置4->5,因爲cur->next!=NULL,因此我們設置cur->right->next=cur->next->left,也就是2的右子樹5指向3的左子樹6,然後cur=3,繼續操作。
代碼如下:
/**
* 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 == NULL) return;
TreeLinkNode *cur;
while(root->left != NULL){
cur = root;
while(cur != NULL){
cur->left->next = cur->right;
if(cur->next != NULL)
cur->right->next = cur->next->left;
cur = cur->next;
}
root = root->left;
}
}
};