[LeetCode] Populating Next Right Pointers in Each Node

最新推荐文章于 2021-09-24 12:05:40 发布

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最新推荐文章于 2021-09-24 12:05:40 发布

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原文链接：http://www.cnblogs.com/vincently/p/4231719.html

本文介绍了一种在完美二叉树中填充每个节点的next指针的方法，使其指向右侧相邻节点。提供了两种解决方案，一种使用递归，另一种通过维护两个指针实现。两种方法均实现了常数级额外空间复杂度。

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,

After calling your function, the tree should look like:

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

思路:前序的递归思路。时间复杂度O(n),,空间复杂度O(1)

 1 /**
 2  * Definition for binary tree with next pointer.
 3  * struct TreeLinkNode {
 4  *  int val;
 5  *  TreeLinkNode *left, *right, *next;
 6  *  TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
 7  * };
 8  */
 9 class Solution {
10 public:
11     void connect(TreeLinkNode *root) {
12        if (root == NULL) return;
13        if (root->left) {
14             root->left->next = root->right;
15             if (root->next)
16                 root->right->next = root->next->left;
17        }
18             
19         connect(root->left);
20         connect(root->right);
21     }
22 };

思路二：维护两个指针。

 1 /**
 2  * Definition for binary tree with next pointer.
 3  * struct TreeLinkNode {
 4  *  int val;
 5  *  TreeLinkNode *left, *right, *next;
 6  *  TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
 7  * };
 8  */
 9 class Solution {
10 public:
11     void connect(TreeLinkNode *root) {
12        if (root == NULL) return;
13        TreeLinkNode *pre = root;
14        TreeLinkNode *cur = NULL;
15        while (pre->left) {
16            cur = pre;
17            while (cur) {
18                cur->left->next = cur->right;
19                if (cur->next) cur->right->next = cur->next->left;
20                cur = cur->next;
21            }
22            
23            pre = pre->left;
24        }
25     }
26 };

转载于:https://www.cnblogs.com/vincently/p/4231719.html