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Copy path145.binary-tree-postorder-traversal.cpp
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145.binary-tree-postorder-traversal.cpp
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// Tag: Stack, Tree, Depth-First Search, Binary Tree
// Time: O(N)
// Space: O(H)
// Ref: -
// Note: PostOrder
// 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:
vector<int> postorderTraversal(TreeNode* root) {
vector<int> results;
helper(root, results);
return results;
}
void helper(TreeNode *node, vector<int> &results) {
if (node == nullptr) {
return;
}
helper(node->left, results);
helper(node->right, results);
results.push_back(node->val);
}
};
class Solution {
public:
vector<int> postorderTraversal(TreeNode* root) {
stack<TreeNode *> st;
vector<int> res;
TreeNode *cur = root;
TreeNode *pre = nullptr;
while (cur || !st.empty()) {
while (cur) {
st.push(cur);
cur = cur->left;
}
cur = st.top();
if (cur->right && cur->right != pre) {
cur = cur->right;
} else {
st.pop();
res.push_back(cur->val);
pre = cur;
cur = nullptr;
}
}
return res;
}
};