Q:
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

My solution

/**
 * 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 == nullptr) {
            return;
        }
        if(root->left != nullptr)
        {
            root->left->next = root->right;
        }
        if(root->right != nullptr && root->next != nullptr)
        {
            root->right->next = root->next->left;
        }
        connect(root->left);
        connect(root->right);
    }
};

這題初看，肯定會用BFS去解決(實際上是可以的),但是其實用DFS會更加方便。該解決方法的核心就是要注意到：root->right->next = root->next->left;。

如果有任何疑問，歡迎留言，或者給我發郵件(liuhiter@gmail.com)。另外無恥求光顧本渣的gayhub leetcode項目。