29. Populating Next Right Pointers in Each Node && Populating Next Right Pointers in Each Node II

Populating Next Right Pointers in Each Node

OJ: https://oj.leetcode.com/problems/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 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

思想： 常量空间要求，决定了不能使用递归。满二叉树，简单循环，每次修改一层即可。

/**
 * 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 *p = root;
        while(p && p->left) {
            p->left->next = p->right;
            TreeLinkNode *pre = p, *cur = p->next;
            while(cur) {
                pre->right->next = cur->left;
                cur->left->next = cur->right;
                pre = cur;
                cur = cur->next;
            }
            p = p->left;
        }
    }
};

Populating Next Right Pointers in Each Node II

OJ:https://oj.leetcode.com/problems/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

思想同上： 但是下一层最开始结点和连接过程中链表的第一个结点不易确定，所以需要设定两个变量来保存。

/**
 * 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 *curListNode, *startNode, *curNode;
        startNode = root;
        while(startNode) {
            curNode = startNode;
            startNode = curListNode = NULL;
            while(curNode) {
                if(curNode->left) {
                    if(startNode == NULL) startNode = curNode->left;
                    if(curListNode == NULL) curListNode = curNode->left;
                    else { curListNode->next = curNode->left; curListNode = curListNode->next; }
                }
                if(curNode->right) {
                    if(startNode == NULL) startNode =curNode->right;
                    if(curListNode == NULL) curListNode = curNode->right;
                    else { curListNode->next = curNode->right; curListNode = curListNode->next; }
                }
                curNode = curNode->next;
            }
        }
    }
};