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
第一眼觉得,诶,用层序遍历不完了吗,但一看只能用常量空间来做,用队列来层序遍历肯定不行的。
还是想了好一会才想到,我们的构造是从上往下的,这一层连结好了,我们当前节点为 pre,那它的兄弟不就是pre->next吗,所以下一层的连接需要两步,一步是pre的左孩子连到右孩子去,一步是pre的右孩子连到 pre->next->left左孩子去,然后依次对pre的兄弟节点处理就完了。
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) { while(root) { TreeLinkNode* pre=root; TreeLinkNode* sib; while(pre) { if(pre->left) pre->left->next=pre->right; sib=pre->next; if ( sib && pre->right) pre->right->next=sib->left; pre=sib; } root=root->left; } } };