LeetCode - 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

http://oj.leetcode.com/problems/populating-next-right-pointers-in-each-node/

Solution:

Recursion. It's always useful in tree problem. Since it is perfect binary tree (ie, all leaves are at the same level, and every parent has two children), we know that every left child has a right brother. For right child, check if the root has a next node.

https://github.com/starcroce/leetcode/blob/master/populating_next_right_pointers_in_each_node.cpp

// 128 ms for 14 test cases
// 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.
/**
 * 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) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        if(root == NULL) {
            return;
        }
        if(root->left != NULL) {
            root->left->next = root->right;
        }
        if(root->right != NULL) {x
            if(root->next != NULL) {
                root->right->next = root->next->left;
            }
            else {
                root->right->next = NULL;
            }
        }
        connect(root->left);
        connect(root->right);
    }
};