怎样python二叉树中的右侧指针,相信很多没有经验的人对此束手无策,为此本文总结了问题出现的原因和解决方法,通过这篇文章希望你能解决这个问题。
给定一个完美二叉树,其所有叶子节点都在同一层,每个父节点都有两个子节点。二叉树定义如下:
struct Node { int val; Node *left; Node *right; Node *next;}
填充它的每个 next 指针,让这个指针指向其下一个右侧节点。如果找不到下一个右侧节点,则将 next 指针设置为 NULL
。
初始状态下,所有 next 指针都被设置为 NULL
。
示例:
输入:{"$id":"1","left":{"$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":{"$id":"7","left":null,"next":null,"right":null,"val":7},"val":3},"val":1}
输出:{"$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}解释:给定二叉树如图 A 所示,你的函数应该填充它的每个 next 指针,以指向其下一个右侧节点,如图 B 所示。
解题思路:
1,树问题均可递归
2,先填充一层,然后递归处理左右子树
3,由于是完美二叉树,处理简单:
左子树的next是右子树,????️子树的next是next的左子树
/*// Definition for a Node.class Node {public: int val; Node* left; Node* right; Node* next; Node() {} Node(int _val, Node* _left, Node* _right, Node* _next) { val = _val; left = _left; right = _right; next = _next; }};*/class Solution {public: Node* connect(Node* root) { if(root==NULL || root->left==NULL){ return root; } root->left->next=root->right; if(root->next!=NULL){ root->right->next=root->next->left; } connect( root->left); connect(root->right); return root; }};
给定一个二叉树
struct Node { int val; Node *left; Node *right; Node *next;}
填充它的每个 next 指针,让这个指针指向其下一个右侧节点。如果找不到下一个右侧节点,则将 next 指针设置为 NULL
。
初始状态下,所有 next 指针都被设置为 NULL
。
示例:
输入:{"$id":"1","left":{"$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":{"$id":"5","left":null,"next":null,"right":{"$id":"6","left":null,"next":null,"right":null,"val":7},"val":3},"val":1}输出:{"$id":"1","left":{"$id":"2","left":{"$id":"3","left":null,"next":{"$id":"4","left":null,"next":{"$id":"5","left":null,"next":null,"right":null,"val":7},"right":null,"val":5},"right":null,"val":4},"next":{"$id":"6","left":null,"next":null,"right":{"$ref":"5"},"val":3},"right":{"$ref":"4"},"val":2},"next":null,"right":{"$ref":"6"},"val":1}解释:给定二叉树如图 A 所示,你的函数应该填充它的每个 next 指针,以指向其下一个右侧节点,如图 B 所示。
提示:
你只能使用常量级额外空间。
使用递归解题也符合要求,本题中递归程序占用的栈空间不算做额外的空间复杂度
解题思路:
1,本题与上题的唯一不同是不是完美二叉树
2,因此需要计算next节点是什么:
A,左子树(若非空)
B,右子树(若非空)
C,next的子树
3,左子树的next 节点有两种情况:
A,右子树(若非空)
B,next(或next的next)节点的子树
4,右子树的next节点:next(或next的next)节点的子树
5,由于计算next节点依赖右子树,所以先递归右子树
/*// Definition for a Node.class Node {public: int val; Node* left; Node* right; Node* next; Node() {} Node(int _val, Node* _left, Node* _right, Node* _next) { val = _val; left = _left; right = _right; next = _next; }};*/class Solution {public: Node* connect(Node* root) { if (root==NULL){ return NULL; } if (root->left!=NULL){ if(root->right!=NULL){ root->left->next=root->right; }else{ root->left->next=nextNode(root->next); } } if(root->right!=NULL){ root->right->next=nextNode(root->next); } connect(root->right); connect(root->left); return root; } Node* nextNode(Node* root){ Node* t=root; while(t!=NULL){ if (t->left!=NULL){ return t->left; } if (t->right!=NULL){ return t->right; } t=t->next; } return NULL; }};
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原文链接:https://my.oschina.net/u/4586289/blog/4634777