Populating Next Right Pointers in Each Node题解

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

我的解法：

（1）算法思想：

如果一个结点root非空且有左结点，将左结点指向右结点，递归connect左结点；如果root的next结点非空，则将root的右结点指向root->next的左结点，再递归connect 右结点。

（2）代码如下：

/**
* Definition for binary tree with next pointer.
* struct TreeLinkNode {
*  int val;
*  TreeLinkNode *left, *right, *next;
*  TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
* };
*/
class Solution {
public:
void connect(TreeLinkNode *root) {
if(!root)return;
if(root&&root->left){
root->left->next=root->right;
connect(root->left);
root->right->next=root->next?root->next->left:NULL;
connect(root->right);
}
}
};