116 Populating Next Right Pointers in Each Node

最新推荐文章于 2022-02-28 18:16:11 发布

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本文探讨了如何在完美二叉树中填充每个节点的Next指针，使其指向右侧相邻节点，若无则设为NULL。通过迭代和递归两种方法详细解析了算法实现，特别强调了仅使用常数额外空间的解决方案。

1 题目

You are given a perfect binary tree where all leaves are on the same level, and every parent has two children. The binary tree has the following definition:

struct Node {
  int val;
  Node *left;
  Node *right;
  Node *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.

Example:

Input: {"$id":"1","left":{"$id":"2","left":{"$id":"3","left":null,"next":null,"right":null,"val":4},"next":null,"right":{"$id":"4","left":null,"next":null,"right":null,"val":5},"val":2},"next":null,"right":{"$id":"5","left":{"$id":"6","left":null,"next":null,"right":null,"val":6},"next":null,"right":{"$id":"7","left":null,"next":null,"right":null,"val":7},"val":3},"val":1}

Output: {"$id":"1","left":{"$id":"2","left":{"$id":"3","left":null,"next":{"$id":"4","left":null,"next":{"$id":"5","left":null,"next":{"$id":"6","left":null,"next":null,"right":null,"val":7},"right":null,"val":6},"right":null,"val":5},"right":null,"val":4},"next":{"$id":"7","left":{"$ref":"5"},"next":null,"right":{"$ref":"6"},"val":3},"right":{"$ref":"4"},"val":2},"next":null,"right":{"$ref":"7"},"val":1}

Explanation: Given the above perfect binary tree (Figure A), your function should populate each next pointer to point to its next right node, just like in Figure B.

Note:

You may only use constant extra space.
Recursive approach is fine, implicit stack space does not count as extra space for this problem.

2 尝试解

class Solution {
public:
    Node* connect(Node* root) {
        Connect(root);
        return root;
    }
    void Connect(Node*root){
        if(root == NULL) return;
        Node* cur = root;  //this is the top layer
        Node* pre = NULL;  //this is the bottom layer
        Node* head = NULL;
        while(cur != NULL){
            if(cur->left != NULL){
                if(pre != NULL)     pre->next = cur->left;
                else head = cur->left;  //find start of the bottom layer to recurse
                pre = cur->left;
            }
            if(cur->right != NULL){
                if(pre != NULL)     pre->next = cur->right;
                else head = cur->right;  //find start of the bottom layer to recurse
                pre = cur->right;
            }
            cur = cur->next;
        }
        Connect(head);
    }
};

3 标准解

void connect(TreeLinkNode *root) {
    if (root == NULL) return;
    TreeLinkNode *pre = root;
    TreeLinkNode *cur = NULL;
    while(pre->left) {
        cur = pre;
        while(cur) {
            cur->left->next = cur->right;
            if(cur->next) cur->right->next = cur->next->left;
            cur = cur->next;
        }
        pre = pre->left;
    }
}