117-populating-next-right-pointers-in-each-node-ii.md

117. Populating Next Right Pointers in Each Node II - 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。

---

题目标签：Tree / Depth-first Search

题目链接：LeetCode / LeetCode中国

题解

Language	Runtime	Memory
cpp	20 ms	2 MB

/**
 * Definition for binary tree with next pointer.
 * struct TreeLinkNode {
 *  int val;
 *  TreeLinkNode *left, *right, *next;
 *  TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
 * };
 */
static auto _ = [](){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    return 0;
}();

class Solution {
public:
    TreeLinkNode* findNext(TreeLinkNode *node) {
        while (node->next) {
            node = node->next;
            if (node->left) return node->left;
            if (node->right) return node->right;
        }
        return NULL;
    }

    void connect(TreeLinkNode *root) {
        if (!root) return;
        if (root->left && root->right) {
            root->left->next = root->right;
            root->right->next = findNext(root);
        } else {
            if (root->left) {
                root->left->next = findNext(root);
            }
            if (root->right) {
                root->right->next = findNext(root);
            }
        }
        connect(root->right);
        connect(root->left);
    }
};