Populating Next Right Pointers in Each Node

描述

Given a binary tree

struct TreeLinkNode {
   int val;
   TreeLinkNode *left, *right, *next;
   TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
};

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.

Note:

• You may only use constant extra space.
• You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).

For example,

         1
       /  \
      2    3
     / \  / \
    4  5  6  7

After calling your function, the tree should look like:

         1 -> NULL
       /  \
      2 -> 3 -> NULL
     / \  / \
    4->5->6->7 -> NULL

分析

无

代码

Python

// Populating Next Right Pointers in Each Node
// 时间复杂度O(n)，空间复杂度O(logn)
public class Solution {
    public void connect(TreeLinkNode root) {
        connect(root, null);
    }
    private static void connect(TreeLinkNode root, TreeLinkNode sibling) {
        if (root == null) return;
        else root.next = sibling;

        connect(root.left, root.right);
        if (sibling != null) connect(root.right, sibling.left);
        else connect(root.right, null);
    }
}

Java

// Populating Next Right Pointers in Each Node
// 时间复杂度O(n)，空间复杂度O(logn)
class Solution {
public:
    void connect(TreeLinkNode *root) {
        connect(root, NULL);
    }
private:
    void connect(TreeLinkNode *root, TreeLinkNode *sibling) {
        if (root == nullptr)
            return;
        else
            root->next = sibling;

        connect(root->left, root->right);
        if (sibling)
            connect(root->right, sibling->left);
        else
            connect(root->right, nullptr);
    }
};

C++

# Populating Next Right Pointers in Each Node
# Time Complexity O(n), Space Complexity O(logn)
class Solution:
    def connect(self, root):
        self._connect(root, None)

    def _connect(self, root, sibling):
        if not root:
            return
        else:
            root.next = sibling

        self._connect(root.left, root.right)
        if sibling:
            self._connect(root.right, sibling.left)
        else:
            self._connect(root.right, None)

相关题目

• Populating Next Right Pointers in Each Node II