C++ Coding Exercise – Binary Tree Preorder Traversal

Return the preorder traversal of a binary tree’s nodes’ values. The pre-order displays the root/current node’s value first, then recursively calling into its left sub tree and right sub tree respectively.

Recursion

Based on the definition of the pre-order, it is trivial to write a recursion:

/**
 * 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:
    void preorderTraversal(TreeNode* root, vector<int> &r) {
        if (root == NULL) {
            return;
        }
        r.push_back(root->val);
        preorderTraversal(root->left, r);
        preorderTraversal(root->right, r);
    }

    vector<int> preorderTraversal(TreeNode* root) {
        vector<int> r;
        preorderTraversal(root, r);
        return r;
    }
};

Without helper function, you can do it straightforward by appending vector to the vector:

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

Please note that it is easy by rearranging the order of r.push_back(root->val); in order to do post-order, in-order traversal.

Using Stack

The iterative approach would be using a stack (First In First Out), so we push the right sub tree first as they are handled later.

class Solution {
public:
    vector<int> preorderTraversal(TreeNode* root) {
        vector<int> r;
        stack<TreeNode*> st;
        if (root) {
            st.push(root);
        }
        while (!st.empty()) {
            auto p = st.top();
            st.pop();
            r.push_back(p->val);
            if (p->right) {
                st.push(p->right);
            }
            if (p->left) {
                st.push(p->left);
            }
        }
        return r;
    }
};

This is the case that using stack to emulate the recursion.

–EOF (The Ultimate Computing & Technology Blog) —