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【LeetCode】Populating Next Right Pointers in Each Node

2024-07-03 19:24:49   224人阅读

Populating Next Right Pointers in Each Node

 

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


先不考虑空间复杂度限制

从图例就可以得到解法，就是层次遍历。
每一层的前一个节点next指向后一个节点，层次之间不相连。
也就是说，
如果当前加入的节点cur是左子结点，那么需要判断一下，该节点是不是新一层的第一个节点。
  若是：上一个节点pre的next不需要指向当前节点
  若否：上一个节点pre的next需要指向当前节点
如果当前加入的节点cur是右子结点，那么不需要判断，上一个节点pre的next需要指向当前节点。
class Solution {
public:
    queue<TreeLinkNode*> q;
    void connect(TreeLinkNode *root) 
    {
        if(root == NULL)
            return;

        int ind_count = 0;
        int pow_count = 1;

        q.push(root);
        ind_count ++;
        TreeLinkNode *cur = root;
        TreeLinkNode *pre = root;
        TreeLinkNode *temp = root;

        while(!q.empty())
        {
            temp = q.front();
            q.pop();

            if(temp->left)
            {
                q.push(temp->left);
                pre = cur;
                cur = temp->left;

                ind_count ++;

                if(ind_count != pow(2.0, pow_count))
                {
                    pre->next = cur;
                }
                else
                {
                    pow_count ++;
                }
            }
            if(temp->right)
            {
                q.push(temp->right);
                pre = cur;
                cur = temp->right;

                ind_count ++;

                pre->next = cur;
            }
        }
    }
};

 



接下来我们考虑怎样去掉队列进行层次遍历。
首先我们需要一个指示器，告诉我们每一层的开始，然后在遍历该层的时候将下一层的next进行连接。（遍历依赖于上一层建立好的next）
想法参考Discussion后自己动手实现
class Solution {
public:

    void connect(TreeLinkNode *root) 
    {
        TreeLinkNode* leftWall = root;
        TreeLinkNode* cur;

        while(leftWall != NULL)
        {
            cur = leftWall;
            
            while(cur != NULL)
            {
                if(cur->left != NULL)
                    cur->left->next = cur->right;

                if(cur->right != NULL && cur->next != NULL)
                    cur->right->next = cur->next->left;

                cur = cur->next;
            }

            leftWall = leftWall->left;
        }
    }
 };


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