leetcode-Populating Next Right Pointers in Each Node

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题目描述

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

解题思路

就是将同层的节点连接起来，最后一个next置空。我们可以利用树的next特性，将某一层的下一层节点的的next特性全部设置好，之后只要从下一层的left节点进入下一行，就可以将下下一层的next特性都设置好。代码如下：

/*** 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;root->next=NULL;while(root){TreeLinkNode *node=root;while(node&&node->left){node->left->next=node->right;node->right->next=node->next==NULL?NULL:node->next->left;node=node->next;//指向本层的下一个节点}root=root->left;//指向下层的第一个节点}}
};

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