题目: 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 binar…

本质上是二叉树的层次遍历,遍历层次的过程当中把next指针加上去. ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ 和问题"Populating Next Right Pointers in Each Node"类似. 如果给定的树是任意的二叉树,你先前的方法还能工作吗? 笔记: 你只能用常量的辅助空间. 例如给定的是羡慕的二叉树, 1 / \ 2 3 / \ \ 4…

[LeetCode]117. Populating Next Right Pointers in Each Node II 解题报告(Python) 标签: LeetCode 题目地址:https://leetcode.com/problems/populating-next-right-pointers-in-each-node-ii/description/ 题目描述: Follow up for problem "Populating Next Right Pointers in Each…

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…

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…

题目链接:Populating Next Right Pointers in Each Node II | LeetCode OJ 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 u…

Populating Next Right Pointers in Each Node OJ: https://oj.leetcode.com/problems/populating-next-right-pointers-in-each-node/ Given a binary tree struct TreeLinkNode { TreeLinkNode *left; TreeLinkNode *right; TreeLinkNode *next; } Populate each next…

Populating Next Right Pointers in Each Node Total Accepted: 72323 Total Submissions: 199207 Difficulty: Medium Given a binary tree struct TreeLinkNode { TreeLinkNode *left; TreeLinkNode *right; TreeLinkNode *next; } Populate each next pointer to poin…

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…

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…