[LeetCode] Populating Next RIght Pointer in Each Node II

最新推荐文章于 2019-06-10 19:43:56 发布

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本文讨论了如何在给定的复杂二叉树中填充每个节点的下一个指针，确保从右到左的链接顺序。通过实例分析，展示了在不同结构的二叉树上应用解决方案的有效性和灵活性。

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Follow up for problem "Populating Next Right Pointers in Each Node".

What if the given tree could be any binary tree? Would your previous solution still work?

Note:

You may only use constant extra space.

For example,
Given the following binary tree,

After calling your function, the tree should look like:

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

Key: consider the case when the next pointer has no child. Must find the following next pointers until one has a child. Populate from the right.

/**
 * 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) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        if(root==NULL || (root->left==NULL && root->right==NULL)) 
            return;
        
        if(root->left!=NULL && root->right!=NULL)
            root->left->next=root->right;
        
        TreeLinkNode *rootLastChild;
        if(root->right!=NULL)
            rootLastChild=root->right;
        else
            rootLastChild=root->left;
        
        TreeLinkNode *nextChild=NULL;
        
        TreeLinkNode *rootNext=root->next;
        while(rootNext!=NULL)
        {
            if(rootNext->left!=NULL)
                nextChild=rootNext->left;
            else if(rootNext->right!=NULL)
                nextChild=rootNext->right;

            if(nextChild != NULL)
                break;

            rootNext=rootNext->next;
        }
        
        if(nextChild != NULL)
        {
            rootLastChild->next=nextChild;
        }
            
        connect(root->right);
        connect(root->left);
    }
};