Populating Next Right Pointers in Each Node II

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.

Example:

Given the following binary tree,

     1
   /  \
  2    3
 / \    \
4   5    7

After calling your function, the tree should look like:

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

 

Approach #1: Java.

/**
 * Definition for binary tree with next pointer.
 * public class TreeLinkNode {
 *     int val;
 *     TreeLinkNode left, right, next;
 *     TreeLinkNode(int x) { val = x; }
 * }
 */
public class Solution {
    public void connect(TreeLinkNode root) {
        TreeLinkNode dummyHead = new TreeLinkNode(0);
        TreeLinkNode pre = dummyHead;
        while (root != null) {
            if (root.left != null) {
                pre.next = root.left;
                pre = pre.next;
            }
            if (root.right != null) {
                pre.next = root.right;
                pre = pre.next;
            }
            root = root.next;
            if (root != null) {
                pre = dummyHead;
                root = dummyHead.next;
                dummyHead.next = null;
            }
        }
    }
}

　　

Appraoch #2: Python.

# Definition for binary tree with next pointer.
# class TreeLinkNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
#         self.next = None

class Solution:
    # @param root, a tree link node
    # @return nothing
    def connect(self, root):
        tail = dummy = TreeLinkNode(0)
        while node:
            tail.next = node.left
            if tail.next:
                tail = tail.next
            tail.next = node.right
            if tail.next:
                tail = tail.next
            node = node.next
            if not node:
                tail = dummy
                node = dummy.next