116. Populating Next Right Pointers in Each Node

本文介绍了一种算法问题的两种解决方案:一种是利用队列进行层级遍历的方法;另一种是采用递归的方式,通过左右子节点间的连接来完成。这两种方法均可使二叉树的每个节点指向其右侧相邻节点。

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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,

    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

解法一,使用队列,时间复杂度O(N):

/**
 * 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) {
        if(root == NULL)
            return ;

        std::queue<TreeLinkNode*> que;
        que.push(root);

        while(!que.empty()){
            const int size = que.size();
            for(int i=0; i<size; ++i){
                TreeLinkNode* tmp = que.front();
                que.pop();

                if(!que.empty() && i != size-1)
                    tmp->next = que.front();
                //default tmp1->next = NULL;

                if(tmp->left != NULL)
                    que.push(tmp->left);
                if(tmp->right != NULL)
                    que.push(tmp->right);
            }
        }
    }
};

解法二,递归:

class Solution {
public:
    void connect(TreeLinkNode *root) {
        if(root == NULL)
            return ;
        if(root->left != NULL){
            root->left->next = root->right; //perfect binary tree
            if(root->next != NULL)
                root->right->next = root->next->left;
        }
        connect(root->left);
        connect(root->right);
    }
};

这道题一开始我使用递归,我一直想借助后续遍历的思路。事实证明,不要死搬硬套!要针对实际情况递归。

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