145. Binary Tree Postorder Traversal

/*
Given a binary tree, return the postorder traversal of its nodes' values.

For example:
Given binary tree {1,#,2,3},
1
\
2
/
3
return [3,2,1].

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

使用迭代实现二叉树的后序遍历

*/

/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<int> postorderTraversal(TreeNode* root) {
vector<int> res;
stack<TreeNode*> st;
TreeNode* lastVisit=NULL;
TreeNode* ptr=root;
while(ptr || !st.empty())
{
if(ptr)
{
st.push(ptr);
ptr=ptr->left;
}
else
{
TreeNode* topNode=st.top();
if(topNode->right && topNode->right!=lastVisit)
ptr=topNode->right;
else
{
res.push_back(topNode->val);
lastVisit=topNode;
st.pop();
}
}
}
return res;
}
vector<int> postorderTraversal2(TreeNode* root) {
vector<int> res;
stack<TreeNode*> st;
if(root) st.push(root);
while(!st.empty())
{
TreeNode* topNode=st.top();
st.pop();
res.push_back(topNode->val);
if(topNode->left) st.push(topNode->left);
if(topNode->right) st.push(topNode->right);
}
reverse(res.begin(),res.end());
return res;
}
};