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leetcode Populating Next Right Pointers in Each Node

时间:2014-12-01 00:48:40      阅读:166      评论:0      收藏:0      [点我收藏+]

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看图就知道想要做什么事了。

Given a binary tree

    struct TreeLinkNode {
      TreeLinkNode *left;
      TreeLinkNode *right;
      TreeLinkNode *next;
    }

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
初始所有人的next都是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).
要求常数额外空间。
 
一个递归就搞定了,就是递归让每一个节点他的左右子树通过next链接,直至到最后一层,然后递归左右节点,继续让他们的左右子树通过next链接。
/**
 * 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:

//每一层把left的right的next设置为right的left
void lr2rl(TreeLinkNode *root)
{
    if (!root) return;
    TreeLinkNode *lr = root -> left, *rl = root -> right;
    
    while(lr && rl)
    {
        lr -> next = rl;
        lr = lr -> right;
        rl = rl -> left;
    }
    lr2rl(root -> left);
    lr2rl(root -> right);
}
    void connect(TreeLinkNode *root) 
    {
        lr2rl(root);
    }
};

 

 

leetcode Populating Next Right Pointers in Each Node

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原文地址:http://www.cnblogs.com/higerzhang/p/4134025.html

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