目錄

• 題目描述：
• 示例：
• 解法：

題目描述：

給定一個二叉樹

struct Node {
int val;
Node left;
Node right;
Node *next;
}
填充它的每個 next 指針，讓這個指針指向其下一個右側節點。如果找不到下一個右側節點，則將 next 指針設置為 NULL。

初始狀態下，所有 next 指針都被設置為 NULL。

示例：

輸入：{"$id":"1","left":{"$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}

輸出：{"$id":"1","left":{"$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}

解釋：給定二叉樹如圖 A 所示，你的函數應該填充它的每個 next 指針，以指向其下一個右側節點，如圖 B 所示。
 

提示：

• 你只能使用常量級額外空間。
• 使用遞歸解題也符合要求，本題中遞歸程序占用的棧空間不算做額外的空間復雜度。

解法：

/*
// 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 root;
        }else{
            Node* head = root;
            Node* pre = NULL;
            Node* cur = head;
            while(head){
                cur = head;
                head = NULL;
                pre = NULL;
                while(cur){
                    if(cur->left){
                        if(head == NULL){
                            head = cur->left;
                        }
                        cur->left->next = cur->right;
                        if(pre == NULL){
                            pre = cur->left;
                        }else{
                            pre->next = cur->left;
                            pre = cur->left;
                        }
                    }
                    
                    if(cur->right){
                        if(head == NULL){
                            head = cur->right;
                        }
                        if(pre == NULL){
                            pre = cur->right;
                        }else{
                            pre->next = cur->right;
                            pre = cur->right;
                        }
                    }
                    cur = cur->next;
                }
            }
            return root;  
        }
    }
};
 

leetcode 117. 填充每個節點的下一個右側節點指針 II(Populating Next Right Pointers in Each Node II)