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 = Noneclass 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