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Binary_Tree_Postorder_Traversal.cpp
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121 lines (115 loc) · 3.8 KB
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class Solution
{
public:
/*用栈,用flag来记录是否被访问*/
vector<int> postorderTraversal(TreeNode* root)
{
typedef struct node_flag
{
TreeNode *node;
int flag = 0;
node_flag(TreeNode * val) :node(val) {}
};
stack<node_flag> stack_tree;
vector<int> result;
if (root == nullptr)
return result;
node_flag flag_root(root);
stack_tree.push(flag_root);
while (!stack_tree.empty())
{
node_flag ptr_tree = stack_tree.top();
if (ptr_tree.flag == 1)
result.push_back(ptr_tree.node->val);
stack_tree.pop();
int left_flag = ptr_tree.flag;
if (ptr_tree.flag == 0)
{
ptr_tree.flag = 1;
stack_tree.push(ptr_tree);
}
if (ptr_tree.node->right != nullptr && left_flag == 0)
{
node_flag flag_right(ptr_tree.node->right);
stack_tree.push(flag_right);
}
if (ptr_tree.node->left != nullptr && left_flag == 0)
{
node_flag flag_left(ptr_tree.node->left);
stack_tree.push(flag_left);
}
}
return result;
}
// Morris 后序遍历,时间复杂度 O(n),空间复杂度 O(1)
vector<int> postorderTraversal(TreeNode *root)
{
vector<int> result;
TreeNode dummy(-1);
TreeNode *cur, *prev = nullptr;
std::function < void(const TreeNode*)> visit =
[&result](const TreeNode *node)
{
result.push_back(node->val);
};
dummy.left = root;
cur = &dummy;
while (cur != nullptr)
{
if (cur->left == nullptr)
{
prev = cur; /* 必须要有 */
cur = cur->right;
}
else
{
TreeNode *node = cur->left;
while (node->right != nullptr && node->right != cur)
node = node->right;
if (node->right == nullptr) /* 还没线索化,则建立线索 */
{
node->right = cur;
prev = cur; /* 必须要有 */
cur = cur->left;
}
else /* 已经线索化,则访问节点,并删除线索 */
{
visit_reverse(cur->left, prev, visit);
prev->right = nullptr;
prev = cur; /* 必须要有 */
cur = cur->right;
}
}
}
return result;
}
private:
// 逆转路径
static void reverse(TreeNode *from, TreeNode *to)
{
TreeNode *x = from, *y = from->right, *z;
if (from == to) return;
while (x != to)
{
z = y->right;
y->right = x;
x = y;
y = z;
}
}
// 访问逆转后的路径上的所有结点
static void visit_reverse(TreeNode* from, TreeNode *to,
std::function< void(const TreeNode*) >& visit)
{
TreeNode *p = to;
reverse(from, to);
while (true)
{
visit(p);
if (p == from)
break;
p = p->right;
}
reverse(to, from);
}
};