LeetCode #94: Binary Tree Inorder Traversal (C/C++).

easy

source: https://leetcode.com/problems/binary-tree-inorder-traversal/
C/C++ Solution to LeetCode problem 94. Binary Tree Inorder Traversal.

Problem

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

Examples

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Example 1:

Input: root = [1,null,2,3]
Output: [1,3,2]

Example 2:

Input: root = []
Output: []

Example 3:

Input: root = [1]
Output: [1]

Constraints

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• The number of 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?

Solution

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Two solutions:

1. Recursively:
   • First call recursively for the left node.
   • Add the value of the node to the result.
   • Call recursively for the right node.
2. Iterative:
   • Using an stack:
   • Push the root node.
   • While the stack is not empty:
     • Get the top node.
       • Push its left node if it hasn’t been visited.
       • Mark the node as visited (for this case, we will use a set to track the visited nodes).
       • Continue until no left nodes are available.
     • Pop the node and add its value to the result.
     • If the node has a right node, push it to the stack.

Solution 1

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/**
 * 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 {
  void  traverse(TreeNode* node, vector<int>& result) {
    if (node == nullptr)
      return;

    traverse(node->left, result);
    result.push_back(node->val);
    traverse(node->right, result);
  }

public:
  vector<int> inorderTraversal(TreeNode* root) {
    vector<int> result;
    traverse(root, result);
    return result;
  }
};

Solution 2

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/**
 * 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> inorderTraversal(TreeNode* root) {
    if (!root)
      return {};

    vector<int> result;
    stack<TreeNode*> st;
    set<TreeNode*> visited;

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

      if (visited.find(t) == visited.end() && t->left) {
        st.push(t->left);
        visited.insert(t);
        continue;
      }

      st.pop();
      result.push_back(t->val);
      if (t->right)
        st.push(t->right);
    }

    return result;
  }
};