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145. Binary Tree Postorder Traversal.cpp
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145. Binary Tree Postorder Traversal.cpp
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/*
Given the root of a binary tree, return the postorder traversal of its nodes' values.
Example 1:
Input: root = [1,null,2,3]
Output: [3,2,1]
Example 2:
Input: root = []
Output: []
Example 3:
Input: root = [1]
Output: [1]
Constraints:
The number of the nodes in the tree is in the range [0, 100].
-100 <= Node.val <= 100
Follow up: Recursive solution is trivial, could you do it iteratively?
*/
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
//1. Recursion
// void postOrder(TreeNode* root, vector<int> &ans){
// if(root == NULL){
// return;
// }
// postOrder(root->left,ans);
// postOrder(root->right,ans);
// ans.push_back(root->val);
// }
vector<int> postorderTraversal(TreeNode* root) {
// vector<int> ans;
// postOrder(root,ans);
// return ans;
//2. Iterative
if(!root)
return {};
vector<int> ans;
stack<TreeNode*> s1;
stack<TreeNode*> s2;
s1.push(root);
while(s1.size()){
auto node = s1.top();
s1.pop();
s2.push(node);
if(node->left) s1.push(node->left);
if(node->right) s1.push(node->right);
}
while(s2.size())
ans.push_back(s2.top()->val), s2.pop();
return ans ;
}
};