Leetcode: 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

按层次遍历可以继续用，不过不是constant extra space.

/**
 * 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) {
        if (root == NULL) return;  
          
        TreeLinkNode *cur = NULL;  
        queue<TreeLinkNode*> trees;  
        trees.push(root);  
        trees.push(NULL);  
        while (!trees.empty()) {  
            cur = trees.front();  
            trees.pop();  
            if (cur != NULL) {  
                cur->next = trees.front();  
                if (cur->left != NULL) {  
                    trees.push(cur->left);
                }
                if (cur->right != NULL) {
                    trees.push(cur->right);  
                }  
            }  
            else if (!trees.empty()) {  
                trees.push(NULL);  
            }  
        }  
    }
};

对应I的第二方法：

class Solution {
public:
    void connect(TreeLinkNode *root) {
        if (root == NULL || root->left == NULL && root->right == NULL) {  
            return;  
        }  
        
        connect(root->left);  
        connect(root->right);  
        
        int level = 0;
        while (true) {
            TreeLinkNode *most_right = mostRight(root->left, level);
            TreeLinkNode *most_left = mostLeft(root->right, level);
            if (most_right != NULL && most_left != NULL) {
                most_right->next = most_left;
                ++level;
            }
            else {
                break;
            }
        } 
    }
    
    TreeLinkNode *mostRight(TreeLinkNode* root, int level) {
        if (root == NULL) {
            return NULL;
        }
        else if (level == 0) {
            return root;
        }
        
        TreeLinkNode *right = mostRight(root->right, level - 1);
        if (right == NULL) {
            right = mostRight(root->left, level - 1);
        }
        
        return right;
    }
    
    TreeLinkNode *mostLeft(TreeLinkNode* root, int level) {
        if (root == NULL) {
            return NULL;
        }
        else if (level == 0) {
            return root;
        }
        
        TreeLinkNode *left = mostLeft(root->left, level - 1);
        if (left == NULL) {
            left = mostLeft(root->right, level - 1);
        }
        
        return left;
    }
};

对应I的第三种方法，基本是从 http://blog.csdn.net/fightforyourdream/article/details/16854731 借鉴而来。学习了。

class Solution {
public:
    void connect(TreeLinkNode *root) {
        if (root == NULL || root->left == NULL && root->right == NULL) {  
            return;  
        }  
        
        TreeLinkNode *rnext = root->next;
        TreeLinkNode *next = NULL;
        while (rnext != NULL && next == NULL) {
            if (rnext->left != NULL) {
                next = rnext->left;
            }
            else if (rnext->right != NULL) {
                next = rnext->right;
            }
            rnext = rnext->next;
        }
        
        if (root->left != NULL) {
            if (root->right != NULL) {
                root->left->next = root->right;
            }
            else {
                root->left->next = next;
            }
        }
        if (root->right != NULL) {
            root->right->next = next;
        }
        
        connect(root->right);
        connect(root->left);  
    }
};

===================第二次=================

/**
 * 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) {
        if (root == NULL || root->left == NULL && root->right == NULL) {
            return;
        }
        
        TreeLinkNode* childNext = NULL;
        TreeLinkNode* next = root->next;
        while (childNext == NULL && next != NULL) {
            if (next->left != NULL) {
                childNext = next->left;
            }
            else if (next->right != NULL) {
                childNext = next->right;
            }
            next = next->next;
        }
        
        if (root->left != NULL) {
            root->left->next = root->right ? root->right : childNext;
        }
        if (root->right != NULL) {
            root->right->next = childNext;
        }
        
        connect(root->right);
        connect(root->left);
    }
};