[LeetCode] Populating Next Right Pointers 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
Tree Depth-first Search
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这道题之所以放上来是因为题目中的那句话:You may only use constant extra space
这就意味着,深搜是不能用的,因为递归是需要栈的,因此空间复杂度将是 O(logn)。毫无疑问广搜也不能用,因为队列也是占用空间的,空间占用还高于 O(logn)
难就难在这里,深搜和广搜都不能用,怎么完成树的遍历?
我拿到题目的第一反应便是:用广搜,接着发现广搜不能用,便犯了难。
看了一些提示,有招了:核心仍然是广搜,但是我们可以借用 next 指针,做到不需要队列就能完成广度搜索。
如果当前层所有结点的next 指针已经设置好了,那么据此,下一层所有结点的next指针 也可以依次被设置。
class Solution { public: void connect(TreeLinkNode *root) { if(root == NULL) return; TreeLinkNode* curLayer = root;//the leftmost node of the layer while(curLayer) { TreeLinkNode* curNode = curLayer ; while(curNode && curNode->left) //curNode is not a leaf node { TreeLinkNode* left = curNode->left; TreeLinkNode* right= curNode->right; left->next = right; if(curNode->next) right->next = curNode->next->left; curNode = curNode->next; } curLayer = curLayer->left; } } };