本文为senlie原创,转载请保留此地址:http://blog.csdn.net/zhengsenlie Populating Next Right Pointers in Each Node II Total Accepted: 9695 Total Submissions: 32965 Follow up for problem "Populating Next Right Pointers in Each Node". What if the given tree could…

Follow up for problem "Populating Next Right Pointers in Each Node". What if the given tree could be any binary tree? Would your previous solution still work? Note: You may only use constant extra space. For example,Given the following binary tr…

Populating Next Right Pointers in Each Node II 题解 原创文章,拒绝转载 题目来源:https://leetcode.com/problems/populating-next-right-pointers-in-each-node-ii/description/ Description Follow up for problem "Populating Next Right Pointers in Each Node". What if t…

Populating Next Right Pointers in Each Node II Follow up for problem "Populating Next Right Pointers in Each Node". What if the given tree could be any binary tree? Would your previous solution still work? Note: You may only use constant extra s…

Follow up for problem "Populating Next Right Pointers in Each Node". What if the given tree could be any binary tree? Would your previous solution still work? Note: You may only use constant extra space.   For example,Given the following binary…

Follow up for problem "Populating Next Right Pointers in Each Node". What if the given tree could be any binary tree? Would your previous solution still work? Note: You may only use constant extra space. For example,Given the following binary tr…

Follow up for problem "Populating Next Right Pointers in Each Node". What if the given tree could be any binary tree? Would your previous solution still work? Note: You may only use constant extra space. For example, Given the following binary t…

void connect(TreeLinkNode *root) { while (root) { //每一层循环时重新初始化 TreeLinkNode *prev = nullptr; TreeLinkNode *next = nullptr; //对于每一层 for (; root; root = root->next) { //每一层开始时,记录下一层的起始结点 if (!next)next = root->left ? root->left : root->right; i…

原题地址 二叉树的层次遍历. 对于每一层,依次把各节点连起来即可. 代码: void connect(TreeLinkNode *root) { if (!root) return; queue<TreeLinkNode *> parents; parents.push(root); while (!parents.empty()) { queue<TreeLinkNode *> children; while (!parents.empty()) { TreeLinkNode *…

leetcode 199. Binary Tree Right Side View 这个题实际上就是把每一行最右侧的树打印出来,所以实际上还是一个层次遍历. 依旧利用之前层次遍历的代码,每次大的循环存储的是一行的节点,最后一个节点就是想要的那个节点 class Solution { public: vector<int> rightSideView(TreeNode* root) { vector<int> result; if(root == NULL) return resul…