1、
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
分析:
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。
1 /** 2 * Definition for binary tree with next pointer. 3 * struct TreeLinkNode { 4 * int val; 5 * TreeLinkNode *left, *right, *next; 6 * TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {} 7 * }; 8 */ 9 class Solution { 10 public: 11 void connect(TreeLinkNode *root) { 12 if(root==NULL) 13 return; 14 15 while(root->left){ 16 TreeLinkNode *cur=root; 17 while(cur!=NULL){ 18 cur->left->next=cur->right; 19 if(cur->next!=NULL){ 20 cur->right->next=cur->next->left; 21 } 22 cur=cur->next; 23 } 24 root=root->left; 25 } 26 } 27 };
2、
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
随后题目做了一些更改:不一定是满二叉树。
解法的核心:递推思想 依然不需要改变,依然是依据当前层的next 指针,设置下一层的 next 指针。只是找结点麻烦些,我们定义了两个函数,findNextNodeNextLev用来找(n+1)层的下一个节点,findStartNodeNextLev 用来找下一层的起始节点。
1 class Solution { 2 public: 3 void connect(TreeLinkNode *root) { 4 if(NULL == root) return; 5 TreeLinkNode* start; 6 TreeLinkNode* curNode; 7 TreeLinkNode* nextNode; 8 while(root != NULL){ 9 start = findStartNodeNextLev(root); 10 curNode = start; 11 nextNode = findNextNodeNextLev(root, start); 12 while(nextNode != NULL){ 13 curNode -> next = nextNode; 14 curNode = nextNode; 15 nextNode = findNextNodeNextLev(root, curNode); 16 } 17 root = start; 18 } 19 } 20 private: 21 TreeLinkNode* findNextNodeNextLev(TreeLinkNode* &cur, TreeLinkNode* curNextLev){ 22 if(cur -> left == curNextLev && cur -> right != NULL){ 23 return cur -> right; 24 }else{ 25 while(cur -> next != NULL){ 26 cur = cur -> next; 27 if(cur -> left != NULL && cur -> left != curNextLev) return cur -> left; 28 if(cur -> right != NULL && cur -> right != curNextLev) return cur -> right; 29 } 30 } 31 return NULL; 32 } 33 34 TreeLinkNode* findStartNodeNextLev(TreeLinkNode* node){ 35 if(NULL == node) return NULL; 36 if(node -> left != NULL) return node -> left; 37 return findNextNodeNextLev(node, node -> left); 38 } 39 };