116. Populating Next Right Pointers in Each Node

2019-11-14 09:05:16

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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. You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children). For 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

解法一，使用队列，时间复杂度O(N）：

/** * Definition for binary tree with next pointer. * struct TreeLinkNode { * int val; * TreeLinkNode *left, *right, *next; * TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {} * }; */class Solution {public: void connect(TreeLinkNode *root) { if(root == NULL) return ; std::queue<TreeLinkNode*> que; que.push(root); while(!que.empty()){ const int size = que.size(); for(int i=0; i<size; ++i){ TreeLinkNode* tmp = que.front(); que.pop(); if(!que.empty() && i != size-1) tmp->next = que.front(); //default tmp1->next = NULL; if(tmp->left != NULL) que.push(tmp->left); if(tmp->right != NULL) que.push(tmp->right); } } }};

解法二，递归：

class Solution {public: void connect(TreeLinkNode *root) { if(root == NULL) return ; if(root->left != NULL){ root->left->next = root->right; //perfect binary tree if(root->next != NULL) root->right->next = root->next->left; } connect(root->left); connect(root->right); }};

这道题一开始我使用递归，我一直想借助后续遍历的思路。事实证明，不要死搬硬套！要针对实际情况递归。

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