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116-Populating-Next-Right-Pointers-in-Each-Node

问题描述

You are given a perfect binary tree where all leaves are on the same level, and every parent has two children. The binary tree has the following definition:

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struct Node {
int val;
Node *left;
Node *right;
Node *next;
}

Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL.

Initially, all next pointers are set to NULL.

Follow up:

  • You may only use constant extra space.
  • Recursive approach is fine, you may assume implicit stack space does not count as extra space for this problem.

Example 1:

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Input: root = [1,2,3,4,5,6,7]
Output: [1,#,2,3,#,4,5,6,7,#]
Explanation: Given the above perfect binary tree (Figure A), your function should populate each next pointer to point to its next right node, just like in Figure B. The serialized output is in level order as connected by the next pointers, with '#' signifying the end of each level.

Constraints:

  • The number of nodes in the given tree is less than 4096.
  • -1000 <= node.val <= 1000

解题思路

  • 根结点的next为null;

  • 除了根结点外,左子节点的next为right;

  • 右子节点的next为父节点的兄弟节点(即父节点的next)或者null;

  • 若是父节点的next为null,则右子节点的next为null;

  • 之后进行递归调用,分别对左右子节点进行connect操作。

  • 代码如下:

  • JavaScript

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var connect = function(root) {
if (!root || !root.left) return root;

root.left.next = root.right;
root.right.next = root.next ? root.next.left : null;

connect(root.left);
connect(root.right);

return root;
};
  • Java
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class Solution {
public Node connect(Node root) {
if (root == null || root.left == null) return root;

root.left.next = root.right;
root.right.next = root.next !=null ? root.next.left : null;

connect(root.left);
connect(root.right);

return root;
}
}
  • c++
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class Solution {
public:
Node* connect(Node* root) {
if (root == NULL || root->left == NULL) return root;

root->left->next = root->right;
root->right->next = root->next !=NULL ? root->next->left : NULL;

connect(root->left);
connect(root->right);

return root;
}
};
  • 时间复杂度:O(n)
  • 空间复杂度:O(1)