notes/OJ notes/pages/Leetcode Binary-Tree-Inorder-Traversal.md

Leetcode Binary-Tree-Inorder-Traversal

2022-07-04 15:42

Algorithms:

#algorithm #DFS #DFS_inorder

Data structures:

#DS #binary_tree

Difficulty:

#coding_problem #difficulty-easy

Additional tags:

#leetcode

Revisions:

N/A

Related topics:

tag:#DFS

• Leetcode Binary-Tree-Postorder-Traversal
• Leetcode Binary-Tree-Preorder-Traversal
• Leetcode Insert-Into-a-Binary-Search-Tree
• Leetcode Invert-Binary-Tree
• Leetcode Lowest-Common-Ancestor-Of-a-Binary-Search-Tree
• Leetcode Maximum-Depth-Of-Binary-Tree
• Leetcode Path-Sum
• Leetcode Search-In-a-Binary-Tree
• Leetcode Symmetric-Tree
• Leetcode Validate-Binary-Search-Tree

Links:

• Link to problem

---

Problem

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

Examples

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

• The number of nodes in the tree is in the range [0, 100].
• -100 <= Node.val <= 100

Thoughts

[!summary] This is a #DFS traversal problem

Many of them are same to Leetcode Binary-Tree-Preorder-Traversal

Solution

Recursion

/**
 * 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 inorder(TreeNode *root, vector<int> &answer) {
    if (!root) {
      return;
    }

    inorder(root->left, answer);
    answer.push_back(root->val);
    inorder(root->right, answer);
  }

public:
  vector<int> inorderTraversal(TreeNode *root) {
    // Recursion.
    vector<int> answer;
    inorder(root, answer);
    return answer;
  }
};

Iteration

/**
 * 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) {
    // Iteration
    stack<TreeNode *> pending;
    vector<int> answer;

    while (root || !pending.empty()) {
      // traverse left first.
      while (root) {
        pending.push(root);
        root = root->left;
      }

      root = pending.top();
      pending.pop();
      answer.push_back(root->val);
      root = root->right;
    }

    return answer;
  }
};