144. Binary Tree Preorder Traversal

---

My Solution

Solution 1: DFS (recursion)

The Key Idea for Solving This Coding Question

C++ Code

/**
 * 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> preorderTraversal(TreeNode *root) {
        dfs(root);
        return preorder;
    }

private:
    vector<int> preorder;

    void dfs(TreeNode *root) {
        if (root == nullptr) {
            return;
        }

        preorder.push_back(root->val);
        dfs(root->left);
        dfs(root->right);
    }
};

Time Complexity

$O(n)$

$n$ is the number of nodes in the binary tree.

Space Complexity

$O(H)$

$H$ is the height of the binary tree.

Solution 2: DFS (iteration, stack)

The Key Idea for Solving This Coding Question

C++ Code 1

/**
 * 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> preorderTraversal(TreeNode *root) {
        stack<TreeNode *> st;
        vector<int> preorder;

        while (!st.empty() || root != nullptr) {
            if (root != nullptr) {
                st.push(root);
                preorder.push_back(root->val);
                root = root->left;
            } else {
                root = st.top();
                st.pop();

                root = root->right;
            }
        }

        return preorder;
    }
};

C++ Code 2

/**
 * 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> preorderTraversal(TreeNode *root) {
        if (root == nullptr) {
            return {};
        }

        stack<TreeNode *> st;
        vector<int> preorder;

        st.push(root);
        while (!st.empty()) {
            TreeNode *node = st.top();
            st.pop();

            preorder.push_back(node->val);
            if (node->right != nullptr) {
                st.push(node->right);
            }
            if (node->left != nullptr) {
                st.push(node->left);
            }
        }

        return preorder;
    }
};

Time Complexity

$O(n)$

$n$ is the number of nodes in the binary tree.

Space Complexity

$O(H)$

$H$ is the height of the binary tree.

Solution 3: Morris Traversal

The Key Idea for Solving This Coding Question

C++ Code

/**
 * 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> preorderTraversal(TreeNode *root) {
        vector<int> preorder;

        while (root != nullptr) {
            if (root->left == nullptr) {
                preorder.push_back(root->val);
                root = root->right;
            } else {
                TreeNode *pre = root->left;
                while (pre->right != nullptr && pre->right != root) {
                    pre = pre->right;
                }

                if (pre->right == nullptr) {
                    preorder.push_back(root->val);
                    pre->right = root;
                    root = root->left;
                } else {
                    pre->right = nullptr;
                    root = root->right;
                }
            }
        }

        return preorder;
    }
};

Time Complexity

$O(n)$

$n$ is the number of nodes in the binary tree.

Space Complexity

$O(1)$

Miscellaneous

94. Binary Tree Inorder Traversal
145. Binary Tree Postorder Traversal