LeetCode #117 - Populating Next Right Pointers in Each Node II

最新推荐文章于 2021-01-29 16:34:55 发布

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本文介绍了一种在二叉树中填充每个节点的next指针的方法，使得每个节点的next指针指向其右侧的相邻节点。该算法仅使用常数额外空间，并允许使用递归方法。通过具体实例展示了算法的效果。

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题目描述：

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.
Recursive approach is fine, implicit stack space does not count as extra space for this problem.

Example:

Given the following binary tree,

After calling your function, the tree should look like:

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

和Populating Next Right Pointers in Each Node类似，但是二叉树不一定是完美二叉树，所以在递归之前要先求出子节点指向的同层节点。

class Solution {
public:
    void connect(TreeLinkNode *root) {
        if(root==NULL) return;
        TreeLinkNode* p=root->next;
        TreeLinkNode* q=NULL;
        while(p!=NULL)
        {
            if(p->left!=NULL)
            {
                q=p->left;
                break;
            }
            else if(p->right!=NULL)
            {
                q=p->right;
                break;
            }
            p=p->next;
        }
        
        if(root->left!=NULL&&root->right!=NULL)
        {
            root->left->next=root->right;
            root->right->next=q;
        }
        else if(root->left!=NULL&&root->right==NULL) root->left->next=q;
        else if(root->left==NULL&&root->right!=NULL) root->right->next=q;
        connect(root->right);
        connect(root->left);
    }
};