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…

You are given a perfect binary tree where all leaves are on the same level, and every parent has two children. The binary tree has the following definition: struct Node { int val; Node *left; Node *right; Node *next; } Populate each next pointer to p…

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

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, al…

题目来源 https://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 pointer to point to its next right node. If th…

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, al…

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, al…

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, a…

题目: 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…

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