leetcode:Populating Next Right Pointers in Each Node II (顺序联接二叉树每一层节点)【面试算法题】

leetcode:Populating Next Right Pointers in Each Node II (顺序连接二叉树每一层节点)【面试算法题】

题目:

Follow up for problem "Populating Next Right Pointers in Each Node".

What if the given tree could be any binary tree? Would your previous solution still work?

Note:

  • You may only use constant extra space.

For example,
Given the following binary tree,

         1
       /  \
      2    3
     / \    \
    4   5    7

After calling your function, the tree should look like:

         1 -> NULL
       /  \
      2 -> 3 -> NULL
     / \    \
    4-> 5 -> 7 -> NULL

题意把二叉树每一层所有的节点顺序链接,只能开辟常数空间。


这题与上道题不同点在于二叉树不是完全二叉树,因此结构差别较大,不能用上道题的方法。

如果当前节点有左右子节点,先把他们链接起来,因为肯定是相邻的。

然后通过root的next指针,找到需要与当前节点子节点连接的后继节点。

因为涉及到root的next指针查询,递归需要先处理完右子树,以便左子树节点next指针的连接。

/**
 * 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:
    void connect(TreeLinkNode *root) {
       if(!root)return;
            TreeLinkNode *temp=root->next,*p;
            if(root->left &&root->right)
            {
                p=root->right;
                root->left->next = root->right;
            }
            else if(root->right)p=root->right;
            else if(root->left)p=root->left;
            else p=NULL;
            if(p)
            {
                while(temp)
                {
                    if(temp->left)
                    {
                        p->next = temp->left;
                        break;
                    }
                    else if(temp->right)
                    {
                        p->next = temp->right;
                        break;
                    }
                    else temp=temp->next;
                }
            }
            connect(root->right);
            connect(root->left);
            
        }
};
// http://blog.csdn.net/havenoidea

 题解目录