Leetcode Populating Next Right Pointers in Each Node

class Solution {
public:
    void connect(TreeLinkNode *root) {
        dfs(root, NULL);
    }
    
    void dfs(TreeLinkNode* root, TreeLinkNode* counter_part_root) {
        if (root == NULL) return;
        root->next = counter_part_root;
        dfs(root->left, root->right);
        dfs(root->right, counter_part_root == NULL ? NULL:counter_part_root->left);
    }
};

貌似做过类似的，水一发

第二轮：

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 to NULL.

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).

 

For example,
Given the following perfect binary tree,

         1
       /  \
      2    3
     / \  / \
    4  5  6  7

 

After calling your function, the tree should look like:

         1 -> NULL
       /  \
      2 -> 3 -> NULL
     / \  / \
    4->5->6->7 -> NULL

/**
 * 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;
        }
        connect(root->left, root->right);
        connect(root->right, NULL);
    }
    
    void connect(TreeLinkNode* root, TreeLinkNode* pairNode) {
        if (root == NULL) {
            return;
        }
        root->next = pairNode;
        
        connect(root->left, root->right);
        connect(root->right, pairNode == NULL ? NULL : pairNode->left);
    }
};

 把用按层遍历的方法用Java写一遍：

/**
 * Definition for binary tree with next pointer.
 * public class TreeLinkNode {
 *     int val;
 *     TreeLinkNode left, right, next;
 *     TreeLinkNode(int x) { val = x; }
 * }
 */
public class Solution {
    public void connect(TreeLinkNode root) {
        TreeLinkNode que = root;
        while (que != null) {
            TreeLinkNode dummyHead = new TreeLinkNode(0);
            TreeLinkNode last = dummyHead, cur = que;
            
            while (cur != null) {
                if (cur.left != null) {
                    last.next = cur.left;
                    last = last.next;
                }
                if (cur.right != null) {
                    last.next = cur.right;
                    last = last.next;
                }
                cur = cur.next;
            }
            que = dummyHead.next;
        }
    }
}

 5.21再写一发C++按层遍历，感觉dfs也挺好

// 00:48
/**
 * 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) {
        TreeLinkNode dummyHead(0);
        TreeLinkNode* clevel = root;
        TreeLinkNode* prev = NULL;
        while (clevel) {
            prev = &dummyHead;
            prev->next = NULL;
            while (clevel) {
                if (clevel->left != NULL) {
                    prev->next = clevel->left;
                    prev = clevel->left;
                }
                if (clevel->right != NULL) {
                    prev->next = clevel->right;
                    prev = clevel->right;
                }
                clevel = clevel->next;
            }
            clevel = dummyHead.next;
        }
    }
};