Populating Next Right Pointers in Each Node

最新推荐文章于 2022-02-28 18:16:11 发布

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本文介绍了一种常数空间解决方案，用于在完美二叉树中填充每个节点的下一个右指针，包括递归方法和层遍历方法。

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Given a binary tree

    struct TreeLinkNode {
      TreeLinkNode *left;
      TreeLinkNode *right;
      TreeLinkNode *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.

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,
Given the following perfect binary tree,

After calling your function, the tree should look like:

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

方法一：递归是容易想到的，但是不保证常数空间！

/**
 * 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) {}
 * };
 */
class Solution {
public:
    void connect(TreeLinkNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(root == NULL || root->left == NULL) return;
        root->left->next = root->right;
        if(root->next != NULL){
            root->right->next = root->next->left;
        }
        connect(root->right);
        connect(root->left);
    }
};

124 milli secs

To get to the leaf nodes, a recursive solution needs to be log2(n) calls deep, and each call has a call stack, which takes up memory. This means that a recursive solution isn't constant memory, but O(log(n)) memory.

方法二：一层一层查找

/**
 * 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) {}
 * };
 */
class Solution {
public:
    void connect(TreeLinkNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(root == NULL) return;
        
        TreeLinkNode* lm = root;
        TreeLinkNode* cur = NULL;
        
        while(lm != NULL)
        {
            cur = lm;
            while(cur != NULL)
            {
                if(cur->left != NULL)
                    cur->left->next = cur->right;
                if(cur->next != NULL && cur->right != NULL)
                    cur->right->next = cur->next->left;
                cur = cur->next;
            }
            lm = lm->left;
        }
        
    }
};

108 milli secs，constant extra space