Populating Next Right Pointers in Each Node II

Q：

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

 

 

A：

使用两个while来搞，这样好像逻辑上清楚一点，内循环只处理当前这一层。

/**
 * 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;
        TreeLinkNode* pre = NULL;
        while (cur_node) {
            bool has_next_level = false;
            while (cur_node) {
                if (cur_node->left) {
                    if (pre != NULL) {
                        pre->next = cur_node->left;
                    } else {
                        next_level_head = cur_node->left;
                    }
                    pre = cur_node->left;
                    has_next_level = true;
                }
                
                if (cur_node->right) {
                    if (pre != NULL) {
                        pre->next = cur_node->right;
                    } else {
                        next_level_head = cur_node->right;                        
                    }
                    pre = cur_node->right;
                    has_next_level = true;
                }
                cur_node = cur_node->next;
            }
            if (!has_next_level) break;
            cur_node = next_level_head;
            pre = NULL;
        }
    }
};