Title: 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.
A: As is inorder traversal, the left node -> the -> Right
(1) a right subtree of the node: the next node is the left node of the right subtree of a node
(2) the node does not have the right subtree is the parent node of the left: the next node is its parent node
(3) The node has no right child, is the right parent node: traversing up, find a node is the left node of its parent node, the next node is the parent node of the node found
/ *
Struct TreeLinkNode {
int Val;
struct TreeLinkNode * left;
struct * TreeLinkNode right;
struct * Next TreeLinkNode; // parent node
TreeLinkNode (int x): val ( x), left (NULL), right (NULL), next ( NULL) {
}
};
* /
class Solution {
public:
TreeLinkNode * GetNext (TreeLinkNode * • pNode)
{
IF (== • pNode nullptr a)
{
return nullptr a;
}
// right subtree has: its next node, this node left and right node of the node
IF (pNode-> = nullptr a right!)
{
• pNode = pNode-> right;
the while (! pNode-> left = nullptr a)
{
= pNode- • pNode> left;
}
return • pNode;
}
the while (pNode-> Next = nullptr a!)
{
// right subtree is not: The left node is the parent node, the next node is its parent node
if ( == pNode- • pNode> next-> left)
{
return pNode-> Next;
}
// right subtree is not: the node is the right node to its parent node, a parent node along the traverse up until you find a parent node left node
the else
{
• pNode = pNode-> Next;
}
}
return nullptr a;
}
};
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