题目:
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:
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。
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(NULL == root) return; TreeLinkNode* curLev; while(root -> left != NULL){ curLev = root; while(curLev != NULL){ curLev -> left -> next = curLev -> right; if(curLev -> next != NULL) curLev -> right -> next = curLev -> next -> left; curLev = curLev -> next; } root = root -> left; } } };
引申:
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:
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。
class Solution { public: void connect(TreeLinkNode *root) { if(NULL == root) return; TreeLinkNode* start; TreeLinkNode* curNode; TreeLinkNode* nextNode; while(root != NULL){ start = findStartNodeNextLev(root); curNode = start; nextNode = findNextNodeNextLev(root, start); while(nextNode != NULL){ curNode -> next = nextNode; curNode = nextNode; nextNode = findNextNodeNextLev(root, curNode); } root = start; } } private: TreeLinkNode* findNextNodeNextLev(TreeLinkNode* &cur, TreeLinkNode* curNextLev){ if(cur -> left == curNextLev && cur -> right != NULL){ return cur -> right; }else{ while(cur -> next != NULL){ cur = cur -> next; if(cur -> left != NULL && cur -> left != curNextLev) return cur -> left; if(cur -> right != NULL && cur -> right != curNextLev) return cur -> right; } } return NULL; } TreeLinkNode* findStartNodeNextLev(TreeLinkNode* node){ if(NULL == node) return NULL; if(node -> left != NULL) return node -> left; return findNextNodeNextLev(node, node -> left); } };