【leetcode】Populating Next Right Pointers in Each Node I & II

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

 

注意是完美二叉树，所以

（1）根据题述：左孩子为空，则右孩子一定为空，所以左孩子为空，则return
（2）如果左孩子不为空，则右孩子一定不为空，所以链接左孩子和右孩子即可（左孩子的next赋值为右孩子）

 

C++：

class Solution {
public:
    void connect(TreeLinkNode *root) {
        
        if(root==NULL) return ;
        
        if(root->left&&root->right)
        {
            root->left->next=root->right;
            if(root->next)
                root->right->next=root->next->left;
            else
                root->right->next=NULL;
        }
            
        connect(root->left);
        connect(root->right);
            
        
    }
};

 

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):
        if root is None:return
        if root.left and root.right:
            root.left.next=root.right
            if root.next:
                root.right.next=root.next.left
            else:
                root.right.next=None
        self.connect(root.left)
        self.connect(root.right)

 

 

II：

Follow up for problem "Populating Next Right Pointers in Each Node".

What if the given tree could be any binary tree? Would your previous solution still work?

Note:

You may only use constant extra space.
For 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

class Solution {
public:
    void connect(TreeLinkNode *root) {
        if(root==NULL) return ;

        TreeLinkNode * p=root;
        TreeLinkNode * q=NULL;
        TreeLinkNode * nextNode=NULL;

        while(p)
        {
            if(p->left)
            {
                if(q)
                    q->next=p->left;
                q=p->left;
                if(nextNode==NULL)
                    nextNode=q;
            }

            if(p->right)
            {
                if(q)
                    q->next=p->right;
                q=p->right;
                if(nextNode==NULL)
                    nextNode=q;
            }

            p=p->next;
        }

           connect(nextNode);
    }
};