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117. Populating Next Right Pointers in Each Node II.md

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117. 填充每个节点的下一个右侧节点指针 II

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。

解法一：

//时间复杂度O(n), 空间复杂度O(n)
/*
// 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* getNext(Node* p) {
        while(p && !p->left && !p->right) p = p->next;
        if(p) p = p->left ? p->left : p->right;
        return p;
    }
    Node* connect(Node* root) {
        if(!root || !root->left && !root->right) return root;
        if(root->left) root->left->next = root->right ? root->right : getNext(root->next);
        if(root->right) root->right->next = getNext(root->next);
        connect(root->right);//先右后左，顺序不能反
        connect(root->left);
        return root;
    }
};

解法一：

思路同第116题。区别在于本题中的二叉树不再是完全二叉树，所以对于右子结点的next指针，需要从父结点不断向右寻找，细节见代码。

2019/11/25 14:11