94. Binary Tree Inorder Traversal

Easy

Problem Description

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]
Jump to Solution: Python Java C++ JavaScript C#

Python Solution


# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right

# Recursive solution
def inorderTraversal(root: Optional[TreeNode]) -> List[int]:
    def inorder(node: Optional[TreeNode]):
        if not node:
            return
        inorder(node.left)
        result.append(node.val)
        inorder(node.right)
    
    result = []
    inorder(root)
    return result

# Iterative solution
def inorderTraversal(root: Optional[TreeNode]) -> List[int]:
    result = []
    stack = []
    curr = root
    
    while curr or stack:
        # Reach the leftmost node
        while curr:
            stack.append(curr)
            curr = curr.left
        
        curr = stack.pop()
        result.append(curr.val)
        curr = curr.right
    
    return result

Java Solution


/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    // Recursive solution
    private List result;
    
    public List inorderTraversal(TreeNode root) {
        result = new ArrayList<>();
        inorder(root);
        return result;
    }
    
    private void inorder(TreeNode node) {
        if (node == null) {
            return;
        }
        inorder(node.left);
        result.add(node.val);
        inorder(node.right);
    }
    
    // Iterative solution
    public List inorderTraversalIterative(TreeNode root) {
        List result = new ArrayList<>();
        Stack stack = new Stack<>();
        TreeNode curr = root;
        
        while (curr != null || !stack.isEmpty()) {
            // Reach the leftmost node
            while (curr != null) {
                stack.push(curr);
                curr = curr.left;
            }
            
            curr = stack.pop();
            result.add(curr.val);
            curr = curr.right;
        }
        
        return result;
    }
}

C++ Solution


/**
 * 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 {
private:
    vector result;
    
    void inorder(TreeNode* node) {
        if (!node) {
            return;
        }
        inorder(node->left);
        result.push_back(node->val);
        inorder(node->right);
    }
    
public:
    // Recursive solution
    vector inorderTraversal(TreeNode* root) {
        result.clear();
        inorder(root);
        return result;
    }
    
    // Iterative solution
    vector inorderTraversalIterative(TreeNode* root) {
        vector result;
        stack stack;
        TreeNode* curr = root;
        
        while (curr || !stack.empty()) {
            // Reach the leftmost node
            while (curr) {
                stack.push(curr);
                curr = curr->left;
            }
            
            curr = stack.top();
            stack.pop();
            result.push_back(curr->val);
            curr = curr->right;
        }
        
        return result;
    }
};

JavaScript Solution


/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number[]}
 */
// Recursive solution
var inorderTraversal = function(root) {
    const result = [];
    
    const inorder = (node) => {
        if (!node) {
            return;
        }
        inorder(node.left);
        result.push(node.val);
        inorder(node.right);
    };
    
    inorder(root);
    return result;
};

// Iterative solution
var inorderTraversalIterative = function(root) {
    const result = [];
    const stack = [];
    let curr = root;
    
    while (curr || stack.length) {
        // Reach the leftmost node
        while (curr) {
            stack.push(curr);
            curr = curr.left;
        }
        
        curr = stack.pop();
        result.push(curr.val);
        curr = curr.right;
    }
    
    return result;
};

C# Solution


/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
public class Solution {
    private IList result;
    
    // Recursive solution
    public IList InorderTraversal(TreeNode root) {
        result = new List();
        Inorder(root);
        return result;
    }
    
    private void Inorder(TreeNode node) {
        if (node == null) {
            return;
        }
        Inorder(node.left);
        result.Add(node.val);
        Inorder(node.right);
    }
    
    // Iterative solution
    public IList InorderTraversalIterative(TreeNode root) {
        IList result = new List();
        Stack stack = new Stack();
        TreeNode curr = root;
        
        while (curr != null || stack.Count > 0) {
            // Reach the leftmost node
            while (curr != null) {
                stack.Push(curr);
                curr = curr.left;
            }
            
            curr = stack.Pop();
            result.Add(curr.val);
            curr = curr.right;
        }
        
        return result;
    }
}

Complexity Analysis

Solution Explanation

This solution provides both recursive and iterative approaches:

Key points: