[Leetcode 82] 116 Populating Next Right Pointers In Each Node

Problem:

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 toNULL.

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

Analysis:

The problem is not that hard and a little tricky. Remember when we are at level i, we can process the link relation of level i+1. And connect the left and right children of a node is easy. But how to connect the right node and left node of two different parent nodes? We can use the root->next->left to get the left node and thus finish the connecting process.

Code:

/**
 * 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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (root == NULL || (root->left == NULL && root->right == NULL))
            return ;
            
        root->left->next = root->right;
        if (root->next != NULL)
            root->right->next = root->next->left;
            
        connect(root->left);
        connect(root->right);
    }
};