[LeetCode]Populating Next Right Pointers in Each Node II

最新推荐文章于 2021-09-24 12:05:40 发布

原创最新推荐文章于 2021-09-24 12:05:40 发布 · 390 阅读

0 ·

CC 4.0 BY-SA版权

LeetCode 专栏收录该内容

149 篇文章

订阅专栏

本文介绍了一种解决二叉树中填充每个节点的Next指针的方法，即使对于任意二叉树也能有效工作，并且仅使用常数额外空间。通过构建一个虚拟头节点来连接同一层级的所有节点，实现了一种新颖而高效的解决方案。

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

巧妙地利用一个头链表完成。

/**
 * 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) {
        TreeLinkNode *ChildFrontNode = new TreeLinkNode (0);//子链表头节点
        TreeLinkNode *pre = ChildFrontNode;//记录先前值，方便生成链表
        while(root){
            if(root->left){
                pre->next = root->left;
                pre = pre->next;
            }
            if(root->right){
                pre->next = root->right;
                pre = pre->next;
            }//两个子节点生成链表
            root = root->next; // 根节点指向下一个
            if(root==NULL){
                root = ChildFrontNode->next;
                pre = ChildFrontNode;
                ChildFrontNode->next = NULL; //走到链表尾部，转到下一行ChildFrontNode
            }
        }
    }
};