Given a binary tree, return the postorder traversal of its nodes' values.
Example:
Input:[1,null,2,3]1 \ 2 / 3 Output:[3,2,1]
Follow up: Recursive solution is trivial, could you do it iteratively?
Method 1. Recursive
/**
* 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) {
if (root == NULL)
return {};
vector<int> res;
postorderTraversal(root, res);
return res;
}
void postorderTraversal(TreeNode* root, vector<int>& res) {
if (root == NULL)
return;
if (root->left)
postorderTraversal(root->left, res);
if (root->right)
postorderTraversal(root->right, res);
res.push_back(root->val);
}
};
Method 2.Iteratively
/**
* 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) {
if (root == NULL)
return {};
vector<int> res;
stack<TreeNode*> s{{root}};
TreeNode* head = root;
while(!s.empty()) {
TreeNode *t = s.top();
if ((!t->left && !t->right) || t->left == head || t->right == head) {
res.push_back(t->val);
s.pop();
head = t;
}
else {
if (t->right)
s.push(t->right);
if (t->left)
s.push(t->left);
}
}
return res;
}
};