Binary Tree Inorder Traversal [leetcode] 非递归的三种解法

第一种方法是Morris Traversal

是O(n)时间复杂度，且不需要额外空间的方法。缺点是需要修改树。

通过将叶子节点的right指向其中序后继。

代码如下

vector<int> inorderTraversal(TreeNode *root) {
vector<int> res;
TreeNode * cur = root;
TreeNode * pre = NULL;
while (cur)
{
if (cur->left == NULL)
{
res.push_back(cur->val);
cur = cur->right;
}
else
{
pre = cur->left;
while (pre->right && pre->right != cur) pre = pre->right;

if (pre->right == NULL)
{
pre->right = cur;
cur = cur->left;
}
if (pre->right == cur)
{
pre->right = NULL;
res.push_back(cur->val);
cur = cur->right;
}
}
}
return res;
}

第二种方法利用栈，模拟递归过程

vector<int> inorderTraversal(TreeNode *root) {
vector<int> res;
vector<TreeNode*> stack;
stack.push_back(root);
set<TreeNode*> visited;
while (stack.size())
{
TreeNode * cur = stack.back();
stack.pop_back();
if (!cur) continue;
if (visited.find(cur) == visited.end())//visited for the first time
{
stack.push_back(cur->right);
stack.push_back(cur);
stack.push_back(cur->left);
visited.insert(cur);
}
else//visited for the second time
res.push_back(cur->val);
}
return res;
}

第三种方法来自《数据结构》

将所有左子节点入栈，当没有左子结点时取栈顶元素打印并转移到右子节点

vector<int> inorderTraversal(TreeNode *root) {
vector<int> res;
vector<TreeNode*> stack;
TreeNode* cur = root;
while (stack.size() || cur)
{
if (cur)
{
stack.push_back(cur);
cur = cur->left;
}
else
{
cur = stack.back();
stack.pop_back();
res.push_back(cur->val);
cur = cur->right;
}
}
return res;
}