标签：ons node constant may pop follow nod roo fine
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:
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
解题思路：与前一题思路一致。用三个指针，父指针用于循环遍历。子指针用于子节点的遍历。子头指针用于记录每一层子指针的头。父指针经历两个循环，外循环用于从上往下遍历，内循环用于层内水平遍历。
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-117-Populating Next Right Pointers in Each Node II
标签：ons node constant may pop follow nod roo fine
原文地址：https://www.cnblogs.com/nobodywang/p/10352816.html
