Populating Next Right Pointers in Each Node

 

Q：

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



A：
好吧，姑且算我是做出来了吧。但是这么多的if else真心让我很崩溃阿。
每次连接一层的节点，然后连接下一层的节点时，就借用上一层的链表。逻辑上还是比较简单的。代码写的不漂亮。

/**
 * 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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (!root) return;
        root->next = NULL;
        TreeLinkNode* cur_node = root;
        TreeLinkNode* next_level_head = root->left;
        while (cur_node) {
            if (!cur_node->left) {
                // cur_node = next_level_head;
                // if (next_level_head) next_level_head = next_level_head->left;
                break;
            } else {
                cur_node->left->next = cur_node->right;
                if (!cur_node->right) {
                    break;
                } else {
                    if (cur_node->next) {
                        cur_node->right->next = cur_node->next->left;
                    } else {
                        cur_node->right->next = NULL;
                    }
                }
            }
            if (!cur_node->next) {
                cur_node = next_level_head;
                if (cur_node) next_level_head = cur_node->left;
            } else {
                cur_node = cur_node->next;
            }
        }
    }
};