【Leetcode】【Medium】Populating Next Right Pointers in Each Node

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

解题思路：

建立两个指针，一个指针用于操作父节点，给孩子结点的next赋值；一个指针用于指向每层的首个结点；

当操作结点处理完一层后，继续处理下一层。

代码：

/**
* 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) {
TreeLinkNode *cur = root;
TreeLinkNode *layer_first = root;

if (!root)
return;

while (layer_first->left) {
cur->left->next = cur->right;
if (cur->next) {
cur->right->next = cur->next->left;
cur = cur->next;
} else {
layer_first = layer_first->left;
cur = layer_first;
}
}

return;
}
};