LeetCode-Populating Next Right Pointers in Each Node

题目

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

---

 

分析

首先要注意的是题目中给出的Note，咱们能够假定全部输入的树都是一颗彻底二叉树，既保证全部的非子节点都有左右两个子节点。 在这个假定下，咱们在进行循环操做时能够方便不少。能够先设置好next指针，而后经过next指针访问其兄弟节点。

# Definition for a  binary tree node
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
#         self.next = None

class Solution:
    # @param root, a tree node
    # @return nothing
    def connect(self, root):
        if not root:
            return None
        last = root
        cur = root
        while last.left: # 保证其不是叶节点
            cur = last
            while cur:
                cur.left.next = cur.right
                if cur.next:
                    # 当前节点右节点的下一个指向当前节点下一个的左节点
                    cur.right.next = cur.next.left 
                cur = cur.next
            last = last.left
        return None