Populating Next Right Pointers in Each Node II

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

码好一次性通过~~ yeah~

思路是很明显的，一层一层的找，利用当前层的next来figure out下一层的next

第一次做的时候用了queue来完成的按层遍历，这次才看到了 O(1)的空间复杂度。

重新写了一下，为了看起来思路清楚点，代码写的略长。

/** * 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:    TreeLinkNode* myNextLine(TreeLinkNode *cur){        if(cur -> left) return cur -> left;        if(cur -> right) return cur -> right;                while(cur -> next){            cur = cur -> next;            if(cur -> left) return cur -> left;            if(cur -> right) return cur -> right;        }                return NULL;    }    TreeLinkNode* myleftnext(TreeLinkNode *cur){        if(cur -> right) return cur -> right;                while(cur -> next){            cur = cur -> next;            if(cur -> left) return cur -> left;            if(cur -> right) return cur -> right;        }                return NULL;    }        TreeLinkNode* myrightnext(TreeLinkNode *cur){        while(cur -> next){            cur = cur -> next;            if(cur -> left) return cur -> left;            if(cur -> right) return cur -> right;        }                return NULL;    }        void connect(TreeLinkNode *root) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        if(!root) return;                TreeLinkNode *cur = root;        TreeLinkNode *nextLine = myNextLine(cur);                while(nextLine){            while(cur){                if(cur -> left){                    cur -> left -> next = myleftnext(cur);                }                if(cur -> right){                    cur -> right -> next = myrightnext(cur);                }                                                cur = cur -> next;            }            cur = nextLine;            nextLine = myNextLine(cur);        }            }};