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【LeetCode从零单刷】Populating Next Right Pointers in Each Node

时间:2015-04-13 00:24:58      阅读:252      评论:0      收藏:0      [点我收藏+]

题目:

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“完美二叉树”这一前提条件,即每一层的节点个数为2的整数次幂,且有左子树时必有右子树。每到整数次幂的节点,next 就指向 NULL.

而且,处理1,然后处理2、3,然后处理4、5、6、7……顺序的访问方式,是不是很像数据结构:队列

/**
 * 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从零单刷】Populating Next Right Pointers in Each Node

原文:http://blog.csdn.net/ironyoung/article/details/45015801

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