Leetcode/微软 -- 116. Populating Next Right Pointers in Each Node(Recur + Iterat)

116. Populating Next Right Pointers in Each Node(Recursion+ Iteration)
medium
https://leetcode.com/problems/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

思路1： use recursion,
Complexity: O(n)time  --  visit each node to set up next pointer.
                     O(logN) space -- each node up to logN nodes from ancestor to itself

public 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;//perfect binary tree, no check if root.right is null
            }
        }
        connect(root.left);
        connect(root.right);
    }
}

思路2: Iterative, need to check cur.next is null, eg rightmost node on level
Complexity: O(N)-time O(1)-space

public class Solution {
    public void connect(TreeLinkNode root) {
        if(root == null)return;
        TreeLinkNode level_start=root;
        while(level_start != null){
            TreeLinkNode cur=level_start;
            while(cur != null){
                if(cur.left != null)cur.left.next = cur.right;
                if(cur.right != null && cur.next != null)cur.right.next = cur.next.left;
                cur = cur.next;
            }
            level_start = level_start.left;
        }
    }
}