Populating Next Right Pointers in Each Node

最新推荐文章于 2022-02-28 18:16:11 发布

原创最新推荐文章于 2022-02-28 18:16:11 发布 · 729 阅读

0 ·

CC 4.0 BY-SA版权

C++ 专栏收录该内容

27 篇文章

订阅专栏

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，使其指向右邻居节点。通过递归方式实现，确保了常数额外空间使用，并保持了二叉树的完美特性。

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

/**
 * 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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(root == NULL)
            return;
        TreeLinkNode* lChild = root->left;
        TreeLinkNode* rChild = root->right;
        if(root->left == NULL || root->right == NULL)
            return;
        while(lChild && rChild)
        {
            lChild->next = rChild;
            lChild = lChild->right;
            rChild = rChild->left;
        }       
        connect(root->left);
        connect(root->right);
    }
};