Flatten Binary Tree to Linked List

         1
/ \
2   5
/ \   \
3   4   6

   1
\
2
\
3
\
4
\
5
\
6

利用前序遍历解决的方法实现

void flatten(TreeNode* root) {
if (root == nullptr)
return;
stack<TreeNode*> sta;
sta.push(root);
TreeNode* lastRoot = root;
while (!sta.empty())
{
root = sta.top();
if (lastRoot != root->right)
{
if (lastRoot != root->left) {
if (root->left != nullptr) {
sta.push(root->left);
continue;
}
}
else
{
TreeNode* tempNode = root->right;
root->right = lastRoot;
root->left = nullptr;
TreeNode* endNode = lastRoot;
while (endNode->right != nullptr)
endNode = endNode->right;
endNode->right = tempNode;
}
if (root->right != nullptr) {
sta.push(root->right);
continue;
}
}
lastRoot = root;
sta.pop();
}
}

void flatten(TreeNode *root) {
TreeNode*now = root;
while (now)
{
if (now->left)
{
//Find current node's prenode that links to current node's right subtree
TreeNode* pre = now->left;
while (pre->right)
{
pre = pre->right;
}
pre->right = now->right;
//Use current node's left subtree to replace its right subtree(original right
//subtree is already linked by current node's prenode
now->right = now->left;
now->left = NULL;
}
now = now->right;
}
}