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  • Populating Next Right Pointers in Each Node II

    题目链接

    Populating Next Right Pointers in Each Node II

    注意点

    • 不要访问空结点
    • 不是完美二叉树

    解法

    解法一:递归,DFS。因为不是完美二叉树所以子树有可能残缺,故需要平行扫描父节点同层的节点,找到他们的左右子节点。然后右节点就是找到的节点,左节点就是如果右节点存在就选右节点否则选找到的节点

    /*
    // 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) return NULL;
            Node* p = root->next;
            while (p) 
            {
                if (p->left) 
                {
                    p = p->left;
                    break;
                }
                if (p->right) 
                {
                    p = p->right;
                    break;
                }
                p = p->next;
            }
            if(root->right) root->right->next = p;
            if(root->left) root->left->next = root->right ? root->right : p; 
            connect(root->right);
            connect(root->left);
            return root;
        }
    };
    

    解法二:非递归。和Populating Next Right Pointers in Each Node - LeetCode一模一样

    /*
    // 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) return NULL;
            queue<Node*> q;
            q.push(root);
            while(!q.empty())
            {
                int size = q.size();
                for(int i = 0;i < size;i++)
                {
                    Node* temp = q.front();q.pop();
                    if(i != size-1) temp->next = q.front();
                    if(temp->left) q.push(temp->left);
                    if(temp->right) q.push(temp->right);
                }
            }
            return root;
        }
    };
    

    小结

    • 只要是遍历都有递归和非递归两种写法
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  • 原文地址:https://www.cnblogs.com/multhree/p/10632362.html
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