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楼主 |
发表于 2020-4-8 12:48:04
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相似题型:116. 填充每个节点的下一个右侧节点指针
- 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。
-
- 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。
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- /*
- // Definition for a Node.
- class Node {
- public:
- int val;
- Node* left;
- Node* right;
- Node* next;
- Node() : val(0), left(NULL), right(NULL), next(NULL) {}
- Node(int _val) : val(_val), left(NULL), right(NULL), next(NULL) {}
- 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 root;
- queue<Node*> temp;
- temp.push(root);
- while(!temp.empty()){
- int len = temp.size();
- Node* temp1 = NULL;
- for(int i = 0; i < len; i++){
- Node* node = temp.front();
- temp.pop();
- if(node == NULL) continue;
- node -> next = temp1;
- temp1 = node;
- temp.push(node -> right);
- temp.push(node -> left);
- }
- }
- return root;
-
- }
- };
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注意事项:1.temp.pop();的书写位置在continue前面。 |
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