LeetCode #116 - Populating Next Right Pointers in Each Node-优快云博客

本文介绍了一种在完美二叉树中进行节点链接的方法，通过递归或迭代方式实现相邻节点之间的next指针连接，使得同一层级的节点能够相互链接。文章提供了两种解决方案，并附带详细的代码实现。

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题目描述：

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.
Recursive approach is fine, implicit stack space does not count as extra space for this problem.
You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).

Example:

Given the following perfect binary tree,

After calling your function, the tree should look like:

     1 -> NULL
   /  \
  2 -> 3 -> NULL
 / \  / \
4->5->6->7 -> NULL

将二叉树每一层都从左到右链接起来，而且只能使用常数空间。

递归的解法较为简单一点，对于当前递归的节点，将左节点指向右节点，将右节点指向根节点指向节点的左节点。

class Solution {
public:
    void connect(TreeLinkNode *root) {
        if(root==NULL) return;
        if(root->left!=NULL) 
        {
            root->left->next=root->right;
            if(root->next!=NULL) root->right->next=root->next->left;
        }
        connect(root->left);
        connect(root->right);
    }
};

迭代的解法要复杂一些，对每一层依次连接，但是需要在遍历每一层之前存储该层的起点，从而在遍历完一层之后，可以继续遍历下一层。

class Solution {
public:
    void connect(TreeLinkNode *root) {
        if(root==NULL) return;
        TreeLinkNode* start=root;
        while(start->left!=NULL)
        {
            TreeLinkNode* cur=start;
            while(cur!=NULL)
            {
                cur->left->next=cur->right;
                if(cur->next!=NULL) cur->right->next=cur->next->left;
                cur=cur->next;
            }
            start=start->left;
        }
        return;
    }
};