【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);
    }
};

注意点：

1） list为非完美二叉树，右分支可能为空，因此从right -> left 遍历

2） 从最右分支开始查找，且root没有 left 节点，则找 right 节点

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();
            }
        }
    }
};

注意点：

1） 新增一个Q2队列，保存下一行的全部元素，辅助判断是最后一个元素（Q为空）则置为NULL

2） queue队列实现比递归要好

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);
            }
        }
    }
};

注意点：

对比第二种方法，改进点是每一行结束处，采用NULL作为标志位