Populating Next Right Pointers in Each Node
Total Accepted:6129 Total Submissions:17922 My SubmissionsGiven 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 toNULL
.
Initially, all next pointers are set toNULL
.
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
直接上递归法
void connect(TreeLinkNode *root)
{
if (!root || !root->left) return;
root->left->next = root->right;
con(root->left, root->right);
}
void con(TreeLinkNode *lTree, TreeLinkNode *rTree)
{
if (lTree->left) lTree->left->next = lTree->right;
if (rTree->left) rTree->left->next = rTree->right;
else return;
lTree->right->next = rTree->left;
con(lTree->left, lTree->right);
con(lTree->right, rTree->left);
con(rTree->left, rTree->right);
}
下面参考leetcode论坛上的程序写了个程序
//新知识点:
//重点注意:利用新构造的数据结构
void connect(TreeLinkNode* root)
{
if (!root || !root->left || !root->right)
return;
TreeLinkNode* rightSibling;
TreeLinkNode* p1 = root;
while (p1)
{
rightSibling = p1->next? p1->next->left:NULL;
p1->left->next = p1->right;
p1->right->next = rightSibling;
p1 = p1->next;
}
connect(root->left);
}
有人会觉得使用递归并不符合常数空间的题意,那么可以改成非递归:
void connect(TreeLinkNode *root)
{
TreeLinkNode *nextLev = root? root->left:NULL;
while (nextLev)
{
for ( ; root; root = root->next)
{
TreeLinkNode *rightSibling = root->next? root->next->left:NULL;
root->left->next = root->right;
root->right->next = rightSibling;
}
root = nextLev;
nextLev = nextLev->left;
}
}
//2014-2-17 update
void connect(TreeLinkNode *root)
{
if (!root) return;
TreeLinkNode *next_level = root->left;
while (next_level)
{
root->left->next = root->right;
TreeLinkNode *pre = root->right;
root = root->next;
while (root)
{
pre->next = root->left;
pre = root->left->next = root->right;
root = root->next;
}
root = next_level;
next_level = root->left;
}
}
//2014-2-17 update
void connect(TreeLinkNode *root)
{
if (!root) return;
while (root->left)
{
TreeLinkNode *next_level = root->left;
root->left->next = root->right;
TreeLinkNode *pre=root->right;
for (root=root->next; root; root = root->next)
{
pre->next = root->left;
pre = root->left->next = root->right;
}
root = next_level;
}
}