struct BinaryTreeNode
{
int value;
BinaryTreeNode *Left;
BinaryTreeNode *Right;
BinaryTreeNode *Parent;
};
BinaryTreeNode* GetNext(BinaryTreeNode* pNode)
{
if (pNode == nullptr)
return nullptr;
BinaryTreeNode* pNew = nullptr;
//如果该结点有右子节点,则下一个结点是它的子树最尽头的左子节点
if (pNode->Right != nullptr)
{
BinaryTreeNode* pRight = pNode->Right;
while (pRight->Left != nullptr)
pRight = pRight->Left;
pNew = pRight;
}
//如果没有右子结点,则下一节点是它的父节点中为左子节点的父结点
else if (pNode->Parent != nullptr)
{
BinaryTreeNode* pCurrent = pNode;
BinaryTreeNode* pParent = pNode->Parent;
while (pParent != nullptr&&pCurrent == pParent->Right)
{
pCurrent = pParent;
pParent = pParent->Parent;
}
pNew = pParent;
}
return pNew;
}
二叉树-----求中序遍历的下一个结点
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转载自blog.csdn.net/weixin_39916039/article/details/82149970
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