117. Populating Next Right Pointers in Each Node II

最新推荐文章于 2021-01-29 16:34:55 发布

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最新推荐文章于 2021-01-29 16:34:55 发布

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本文链接：https://blog.youkuaiyun.com/nzcusing/article/details/60209697

本文介绍了一种解决二叉树中填充相邻节点指针的问题的方法，特别针对任何形态的二叉树。通过巧妙利用递归技术，该算法能够在常数额外空间条件下完成任务。

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问题描述
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
```
```
    1 -> NULL
   /  \
  2 -> 3 -> NULL
 / \    \
4-> 5 -> 7 -> NULL
```
解决思路
116其实就是117的一个特例。所以这里直接贴出117的解决办法就好了。主要是利用递归。一层一层连接，每一层都找到开始的节点，然后用这个节点递归。其实这道题主要是情况讨论题，不难。
代码

/**
 * 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) {
        helper(root);
    }
    void helper(TreeLinkNode *root) {
        if (!root)
            return;
        TreeLinkNode *cur = root;
        TreeLinkNode *next_pre = NULL,*next_begin = NULL;

        while(cur != NULL) {
            if (!cur->left && !cur->right) {
                cur = cur->next;
            } else if (cur->left && cur->right) {
                if (!next_pre) {
                    cur->left->next = cur->right;
                    next_pre = cur->right;
                    next_begin =cur->left;
                } else {
                    next_pre->next = cur->left;
                    cur->left->next = cur->right;
                    next_pre = cur->right;
                }
                cur = cur->next;
            } else if (!cur->right) {
                if (!next_pre) {
                    next_pre = cur->left;
                    next_begin = cur->left;
                } else {
                    next_pre->next = cur->left;
                    next_pre = cur->left;
                }
                cur = cur->next;
            } else {
                if (!next_pre) {
                    next_pre = cur->right;
                    next_begin = cur->right;
                } else {
                    next_pre->next = cur->right;
                    next_pre = cur->right;
                }
                cur = cur->next;
            }
        }
        helper(next_begin);
    }
};