LeetCode | Populating Next Right Pointers in Each Node II

题目：

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

思路：

类似/article/1382601.html。只是在寻找子节点的时候需要多一些考虑。使用递归的方法可以使代码更加简介。

代码：

非递归方法：

/**
 * 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 != NULL)
        {
            build(root);
            if(root->left != NULL)
            {
                connect(root->left);
            }
            if(root->right != NULL)
            {
                connect(root->right);
            }
        }
    }
    
    void build(TreeLinkNode * root)
    {
        if(root != NULL)
        {
            TreeLinkNode * parent = root;
            TreeLinkNode * cur = NULL;
            TreeLinkNode * next = NULL;
            int isLeft = 0;
            do
            {
                while(cur == NULL && parent != NULL)
                {
                    if(parent->left != NULL)
                    {
                        isLeft = 1;
                        cur = parent->left;
                        break;
                    }
                    if(parent->right != NULL)
                    {
                        isLeft = 2;
                        cur = parent->right;
                        break;
                    }
                    parent = parent->next;
                }
                if(isLeft == 2 && parent != NULL)
                {
                    parent = parent->next;
                    isLeft = 0;
                }
                while(next == NULL && parent != NULL)
                {
                    if(parent->left != NULL && isLeft != 1)
                    {
                        isLeft = 1;
                        next = parent->left;
                        break;
                    }
                    if(parent->right != NULL)
                    {
                        isLeft = 2;
                        next = parent->right;
                        break;
                    }
                    if(isLeft == 1)
                    {
                        isLeft = 0;
                    }
                    parent = parent->next;
                }
                if(cur != NULL && next != NULL)
                {
                    cur->next = next;
                    cur = next;
                    next = NULL;
                }
            }
            while(parent != NULL);
        }
    }
};

递归方法：

/**
 * 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) {
        while((root = getChildren(root, NULL)) != NULL);
    }
    
    TreeLinkNode* getChildren(TreeLinkNode* node, TreeLinkNode* cur){
        if(node == NULL){
            return NULL;
        }

        TreeLinkNode* firstChild = NULL;
        if(node->left != NULL){
            setCurrentNode(cur, node->left);
            if(firstChild == NULL){
                firstChild = cur;
            }
        }
        
        if(node->right != NULL){
            setCurrentNode(cur, node->right);
            if(firstChild == NULL){
                firstChild = cur;
            }
        }
        
        TreeLinkNode* tmp = getChildren(node->next, cur);
        if(firstChild == NULL){
            firstChild = tmp;
        }
        return firstChild;
    }
    
    void setCurrentNode(TreeLinkNode* &cur, TreeLinkNode* next){
        if(cur == NULL){
            cur = next;
        }
        else{
            cur->next = next;
            cur = next;
        }
    }
};