Leetcode: Binary Tree Preorder Traversal

Given a binary tree, return the preorder traversal of its nodes' values.

For example:

Given binary tree 

{1,#,2,3}

,

1
\
2
/
3

return 

[1,2,3]

.

Note: Recursive solution is trivial, could you do it iteratively?

Recursive:

/**
* Definition for binary tree
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
void preorder(TreeNode* root, vector<int> &path)
{
if(root!=NULL)
{
path.push_back(root->val);
preorder(root->left, path);
preorder(root->right, path);
}
}
vector<int> preorderTraversal(TreeNode *root) {
// IMPORTANT: Please reset any member data you declared, as
// the same Solution instance will be reused for each test case.
vector<int> path;
preorder(root, path);
return path;
}
};

Iterative:

vector<int> preorderTraversal(TreeNode *root) {
// IMPORTANT: Please reset any member data you declared, as
// the same Solution instance will be reused for each test case.
vector<int> path;
stack<TreeNode*> stk;
while(root != NULL || !stk.empty())
{
if(root != NULL)
{
while(root)
{
path.push_back(root->val);
stk.push(root);
root=root->left;
}
}
else{
root = stk.top()->right;
stk.pop();
}
}
return path;
}

vector<int> preorderTraversal(TreeNode *root) {
// IMPORTANT: Please reset any member data you declared, as
// the same Solution instance will be reused for each test case.
vector<int> path;
stack<TreeNode*> stk;
if(root != NULL)
{
stk.push(root);
while(!stk.empty())
{
root = stk.top();
stk.pop();
path.push_back(root->val);
if(root->right)	stk.push(root->right);
if(root->left) stk.push(root->left);
}
}
return path;
}

其他两种非递归的遍历算法：

中序遍历-非递归：

vector<int> InOrderTraversal(TreeNode *root) {
// IMPORTANT: Please reset any member data you declared, as
// the same Solution instance will be reused for each test case.
vector<int> path;
stack<TreeNode*> stk;
while(root != NULL || !stk.empty())
{
while(root != NULL)
{
stk.push(root);
root = root->left;
}
if( !stk.empty())
{
root = stk.top();
stk.pop();
root = root->right;
}
}

return path;
}

后序遍历-非递归：

struct TNode{
//TreeNode* root;
int val;
TNode *left;
TNode *right;
bool isVisited;
TNode(int a):val(a),left(NULL),right(NULL),isVisited(false){};
};

vector<int> PostOrderTraversal(TNode *root) {
// IMPORTANT: Please reset any member data you declared, as
// the same Solution instance will be reused for each test case.
vector<int> path;
stack<TNode*> stk;
if(root != NULL)
{
root->isVisited = false;
stk.push(root);
}
TNode* cur = NULL;
while(!stk.empty())
{
cur = stk.top();
if(cur->isVisited)
{
path.push_back(cur->val);
stk.pop();
}else{
cur->isVisited=true;
if(cur->right)
{
cur->right->isVisited = false;
stk.push(cur->right);
}
if(cur->left)
{
cur->left->isVisited = false;
stk.push(cur->left);
}
}
}
return path;
}