Given a binary tree, return the postorder traversal of its nodes' values.
2017-03-17 14:21
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/**
* Definition for binary tree
* 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> result;
stack<TreeNode*> tempStack;
TreeNode *dummyRootNode = root;
if(dummyRootNode == NULL)
return result;
tempStack.push(dummyRootNode);
while(!tempStack.empty())
{
TreeNode *tempNode = tempStack.top();
tempStack.pop();
result.push_back(tempNode->val);
if(tempNode1->left) tempStack.push(tempNode->left);
if(tempNode1->right) tempStack.push(tempNode->right);
}
reverse(result.begin(),result.end());
return result;
}
};
* Definition for binary tree
* 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> result;
stack<TreeNode*> tempStack;
TreeNode *dummyRootNode = root;
if(dummyRootNode == NULL)
return result;
tempStack.push(dummyRootNode);
while(!tempStack.empty())
{
TreeNode *tempNode = tempStack.top();
tempStack.pop();
result.push_back(tempNode->val);
if(tempNode1->left) tempStack.push(tempNode->left);
if(tempNode1->right) tempStack.push(tempNode->right);
}
reverse(result.begin(),result.end());
return result;
}
};
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