LeetCode题解：Populating Next Right Pointers in Each Node I and II

Populating Next Right Pointers in Each Node

Given a binary tree

    struct TreeLinkNode {
      TreeLinkNode *left;
      TreeLinkNode *right;
      TreeLinkNode *next;
    }

Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set toNULL.

Initially, all next pointers are set to NULL.

Note:

• You may only use constant extra space.
• You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).

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

思路：

对树进行广度优先搜索即可。

题解：

/**
 * 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 == nullptr)
            return;
        
        queue<pair<TreeLinkNode*, int>> traverse;
        traverse.push(make_pair(root, 0));
        while(!traverse.empty())
        {
            auto data = traverse.front();
            traverse.pop();
            
            if (data.first->left != nullptr)
                traverse.push(make_pair(data.first->left, data.second + 1));
            if (data.first->right != nullptr)
                traverse.push(make_pair(data.first->right, data.second + 1));
            
            if (traverse.empty() || traverse.front().second != data.second)
                data.first->next = nullptr;
            else
                data.first->next = traverse.front().first;
        }
    }
};

思路：

刚才没有考虑到空间复杂度要求是O(1)。其实对搜索不可避免用到递归等手法，这样栈的空间复杂度就是O(log N)，N为树深。要利用next指针获得相邻树的最左元素。第二题的思路也是类似，只不过要通过next指针找到下一个有子结点的同层结点，这样的话就必须首先保证树的右边部分的next指针有效。

题解：

/**
 * 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 == nullptr)
            return;
        
        if (root->left != nullptr)
            root->left->next = root->right;
        
        if (root->right != nullptr && root->next != nullptr)
            root->right->next = root->next->left;
        else if (root->right != nullptr)
            root->right->next = nullptr;
            
        connect(root->left);
        connect(root->right);
    }
};

/**
 * 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* node) {
        if (node == nullptr)
            return;
            
        if (node->left && node->right)
            node->left->next = node->right;
        
        TreeLinkNode* child = node->left;
        if (node->right) child = node->right;
        
        if (child != nullptr)
        {
            TreeLinkNode* i = node->next;
            while( i != nullptr && i->left == nullptr && i->right == nullptr)
                i = i->next;
            
            if (i != nullptr)
            {
                if (i->left)
                    child->next = i->left;
                else
                    child->next = i->right;
            }
        }
        
        // This is critical since we need the right hand side
        // next is ready before the left hand side
        if (node->right) connect(node->right);
        if (node->left) connect(node->left);
    }
};