logseq_notes/pages/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_problems #difficulty_easy

Additional tags:

#leetcode

Revisions:

N/A

Related topics:

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;
}
};