【leetcode】Populating Next Right Pointers in Each Node II

Question：

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

Anwser 1：    

/** * 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(NULL == root){            return;        }        TreeLinkNode *cur = root->next;        TreeLinkNode *p = NULL;                while(cur != NULL){     // find last right node (left or right)            if(cur->left) {                p = cur->left;                break;            }            if(cur->right){                p = cur->right;                break;            }            cur = cur->next;        }                if(root->right){            root->right->next = p;        }                if(root->left){            root->left->next = root->right ? root->right : p;        }                connect(root->right);   // from right to left        connect(root->left);    }};

注意點：

1） list為非完美二叉樹，右分支可能為空白，因此從right -> left 遍曆

2） 從最右分支開始尋找，且root沒有 left 節點，則找 right 節點

Anwser 2：     

/** * 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(NULL == root){            return;        }                queue<TreeLinkNode *> Q;    // save one line root(s)        queue<TreeLinkNode *> Q2;   // save next one line root(s), swap with Q        Q.push(root);                while(!Q.empty()){            TreeLinkNode *tmp = Q.front();            Q.pop();                        if(tmp->left) Q2.push(tmp->left);            if(tmp->right) Q2.push(tmp->right);                        if(Q.empty()){                tmp->next = NULL;                queue<TreeLinkNode*> tmpQ = Q;  // swap queue                Q = Q2;                Q2 = tmpQ;            } else {                tmp->next = Q.front();            }        }    }};

注意點：

1） 新增一個Q2隊列，儲存下一行的全部元素，輔助判斷是最後一個元素（Q為空白）則置為NULL

2） queue隊列實現比遞迴要好

Anwser 3： 

/** * 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(NULL == root){            return;        }                queue<TreeLinkNode *> Q;    // save one line root(s)        Q.push(root);        Q.push(NULL);       // flag                while(!Q.empty()){            TreeLinkNode *tmp = Q.front();            Q.pop();                        if(tmp != NULL){        // check flag                if(tmp->left) Q.push(tmp->left);                if(tmp->right) Q.push(tmp->right);                tmp->next = Q.front();            } else {                        if(Q.empty()){      // pop flag = NULL, then check is empty                    break;                }               Q.push(NULL);            }        }    }};

注意點：

對比第二種方法，改進點是每一行結束處，採用NULL作為標誌位