LeetCode OJ 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

思路就是用一个数组记录树的每一层的节点，当同一层有新节点的时候，将旧节点的next指向新节点并将新节点放入到数组中：

class Solution {
public:
    void connect(TreeLinkNode * root) {
        TreeLinkNode * leftWall = root;
        while (leftWall != NULL) {
            TreeLinkNode * across = leftWall;
            while (across != NULL) {
                if (across->left != NULL) {
                    across->left->next = across->right;
                }
                if (across->right != NULL && across->next != NULL) {
                    across->right->next = across->next->left;
                }
                across = across->next;
            }
            leftWall = leftWall->left;
        }
    }
};