【LeetCode】82. Populating Next Right Pointers in Each Node
阿新 • • 發佈:2018-12-12
題目描述(Hard)
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.
- Recursive approach is fine, implicit stack space does not count as extra space for this problem.
- You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).
題目連結
Example 1:
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
演算法分析
提交程式碼:
class Solution { public: void connect(TreeLinkNode *root) { connect(root, NULL); } void connect(TreeLinkNode *root, TreeLinkNode *brother) { if (!root) return; root->next = brother; connect(root->left, root->right); if (brother) connect(root->right, brother->left); else connect(root->right, NULL); } };
提交程式碼:
class Solution {
public:
void connect(TreeLinkNode *root) {
if (!root) return;
while(root)
{
// 下一層的首節點
TreeLinkNode *next = NULL;
// 前驅
TreeLinkNode *prev = NULL;
for (; root; root = root->next)
{
if (!next) next = root->left ? root->left : root->right;
if (root->left)
{
if (prev) prev->next = root->left;
prev = root->left;
}
if (root->right)
{
if (prev) prev->next = root->right;
prev = root->right;
}
}
root = next;
}
}
};