145 Binary Tree Postorder Traversal¶

Given a binary tree, return the postorder traversal of its nodes’ values.

For example: Given binary tree {1,#,2,3},

return [3,2,1].

Note: Recursive solution is trivial, could you do it iteratively?

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Solution 1:¶

class Solution {
public:
    vector<int> postorderTraversal(TreeNode* root) {
        vector<int> ans;
        postorderTraversal(root, ans);
        return ans;
    }

    void postorderTraversal(TreeNode* root, vector<int>& ans) {
        if (!root) return;
        postorderTraversal(root->left, ans);
        postorderTraversal(root->right, ans);
        ans.push_back(root->val);
    }
};

Solution 2:¶

class Solution {
public:
    vector<int> postorderTraversal(TreeNode* root) {
        if (!root) return {};
        vector<int> ans;
        const vector<int> l = postorderTraversal(root->left);
        const vector<int> r = postorderTraversal(root->right);
        ans.insert(ans.end(), l.begin(), l.end());
        ans.insert(ans.end(), r.begin(), r.end());
        ans.push_back(root->val);
        return ans;
    }
};

Solution 3:¶

class Solution {
public:
    vector<int> postorderTraversal(TreeNode* root) {
        if (!root) return {};
        deque<int> ans;
        stack<TreeNode*> s;
        s.push(root);
        while (!s.empty()) {
            TreeNode* n = s.top();
            s.pop();
            ans.push_front(n->val); // O(1)
            if (n->left) s.push(n->left);
            if (n->right) s.push(n->right);
        }
        return vector<int>(ans.begin(), ans.end());
    }
};