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社区首页 >专栏 >Leetcode 116 Populating Next Right Pointers in Each Node

Leetcode 116 Populating Next Right Pointers in Each Node

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triplebee
发布2018-01-12 14:53:51
4120
发布2018-01-12 14:53:51
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Given a binary tree

代码语言:javascript
复制
    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,

代码语言:javascript
复制
         1
       /  \
      2    3
     / \  / \
    4  5  6  7

After calling your function, the tree should look like:

代码语言:javascript
复制
         1 -> NULL
       /  \
      2 -> 3 -> NULL
     / \  / \
    4->5->6->7 -> NULL

将二叉树的每一层节点从左至右连接起来。

可以用层序遍历,也可以深搜。

代码语言:javascript
复制
/**
 * 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) return ;
        if(root->left) 
        {
            root->left->next=root->right;
            connect(root->left);
        }
        if(root->right) 
        {
            if(root->next) root->right->next=root->next->left;
            connect(root->right);
        }
    }
};
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原始发表:2016-10-14 ,如有侵权请联系 cloudcommunity@tencent.com 删除

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