LeetCode-116-Populating Next Right Pointers in Each Node
阿新 • • 發佈:2019-02-05
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算法描述:
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).
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
解題思路:用三個指針,父指針用於循環遍歷。子指針用於子節點的遍歷。子頭指針用於記錄每一層子指針的頭。父指針經歷兩個循環,外循環用於從上往下遍歷,內循環用於層內水平遍歷。
void connect(TreeLinkNode *root) { TreeLinkNode* childHead = nullptr; TreeLinkNode* child = nullptr; while(root!=nullptr){ while(root!=nullptr){ if(root->left!=nullptr){ if(childHead!=nullptr) child->next = root->left; else childHead = root->left; child = root->left; } if(root->right != nullptr){ if(childHead!=nullptr) child->next = root->right; else childHead = root->right; child = root->right; } root = root->next; } root = childHead; childHead = nullptr; child = nullptr; } }
LeetCode-116-Populating Next Right Pointers in Each Node