Prove safety face questions offer the next node binary tree 8

Problem: Given a binary tree and a node which is, find the next node in a preorder traversal order and returns. Note that the node in the tree contains not only the left and right child nodes, the parent node contains a pointer pointing to.

Input: a node pointer in the binary tree

Output: pointer to the next node

Ideas:

When a node has a right child , it is the next node in the right subtree of the left-most node.

When a node is not the right sub-tree , it is the parent of the left son , it's the next node parent node.

When a node is not the right sub-tree , it is the parent of the right son , it is either the next node (retroactive up the parent, until you find a node as its parent node of the left subtree), either as ( If not retroactive, compared nullptr).

Code:

``````/*
int val;
TreeLinkNode(int x) :val(x), left(NULL), right(NULL), next(NULL) {

}
};
*/
class Solution {
public:
{
//if(pNode->next==nullptr)  初始为了只有一个节点，考虑，但其实是多虑了，反而会影响判断。
//    return nullptr;
if(pNode->right!=nullptr)
{
result = pNode->right;
while(result->left!=nullptr)
{
result = result->left;
}
return result;
}
if(pNode->next!=nullptr&&pNode->next->left==pNode)
return pNode->next;
if(pNode->next!=nullptr&&pNode->next->right==pNode)  // 这里不能用else，else包含的情况太多了
{
if(pNode->next!=nullptr)
result = pNode->next;
while(result->next!=nullptr&&result->next->right==result)
{
result= result->next;
}
return result->next;
}
return nullptr;
}
};``````

The above second and third clauses if it may be replaced by the following code.

`````` else if(pNode->m_pParent != nullptr)
{
BinaryTreeNode* pCurrent = pNode;
BinaryTreeNode* pParent = pNode->m_pParent;
while(pParent != nullptr && pCurrent == pParent->m_pRight)  // 向上追溯节点为父节点的左子树的父节点
{
pCurrent = pParent;
pParent = pParent->m_pParent;
}

pNext = pParent;
}``````

Complexity Analysis: time complexity of O (n), the spatial complexity is O (1).

Topic: Another idea: The reduction sequence preorder binary tree . Disadvantages: will increase the space complexity, little introduction here.

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Origin blog.csdn.net/qq_22148493/article/details/104345775
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