【LeetCode從零單刷】Populating Next Right Pointers in Each Node

標籤：leetcode   c++   二叉樹   隊列   

題目：

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, 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

解答：

要利用好“完美二叉樹”這一前提條件，即每一層的節點個數為2的整數次冪，且有左子樹時必有右子樹。每到整數次冪的節點，next 就指向 NULL.

而且，處理1，然後處理2、3，然後處理4、5、6、7……順序的訪問方式，是不是很像資料結構：隊列。

/** * 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;        if(root->left == NULL && root->right == NULL)   return;                queue<TreeLinkNode*> s;        s.push(root);        int i = 0;        int j = 1;        TreeLinkNode *tmp;        while(!s.empty())        {            tmp = s.front();            s.pop();            i++;            if(i == j)            {                i = 0;                j = j*2;                tmp->next = NULL;            }            else            {                tmp->next = s.front();            }                        if(tmp->left)            {                s.push(tmp->left);                s.push(tmp->right);            }                    }    }};

【LeetCode從零單刷】Populating Next Right Pointers in Each Node