leetcode: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?

Subscribe to see which companies asked this question

/**
* 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 {

private:
void pushNodeToStack(TreeNode* root, stack<TreeNode*> &auxStack) {

TreeNode* curNode = root;
while (curNode->left != NULL || curNode->right != NULL)
{
if (curNode->right)
auxStack.push(curNode->right);
if (curNode->left)
auxStack.push(curNode->left);

curNode = auxStack.top();
}
return;
}

public:
vector<int> postorderTraversal(TreeNode* root) {

vector<int> retVtr;
if (root == NULL)
return retVtr;

stack<TreeNode *> auxStack;
auxStack.push(root);
pushNodeToStack(root, auxStack);

while (auxStack.size() > 0)
{
TreeNode *preNode = NULL;
while (auxStack.size() > 0 && (preNode == auxStack.top()->left || preNode == auxStack.top()->right))
{
TreeNode *curNode = auxStack.top();
preNode = curNode;
auxStack.pop();
retVtr.push_back(curNode->val);
}
if (auxStack.size() > 0)
pushNodeToStack(auxStack.top(), auxStack);
}

return retVtr;
}
};