94. Binary Tree Inorder Traversal

https://leetcode.com/problems/binary-tree-inorder-traversal/

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */

/**
 * @param {TreeNode} root
 * @return {number[]}
 */
function inorderTraversal(root) {
  if (!root) {
    return []
  }

  let result = []
  result = result.concat(inorderTraversal(root.left))
  result.push(root.val)
  result = result.concat(inorderTraversal(root.right))
  return result
}

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */

/**
 * @param {TreeNode} root
 * @return {number[]}
 */
function inorderTraversal(root) {
  if (!root) {
    return []
  }

  const result = []
  const stack = []
  let current = root

  while (current || stack.length) {
    while (current) {
      stack.push(current)
      current = current.left
    }
    current = stack.pop()
    result.push(current.val)
    current = current.right
  }

  return result
}

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


class Solution(object):

    def inorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        if not root:
            return []
        result = []
        result.extend(self.inorderTraversal(root.left))
        result.append(root.val)
        result.extend(self.inorderTraversal(root.right))
        return result

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


class Solution(object):

    def inorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        if not root:
            return []
        result = []
        stack = []
        current = root
        while current or stack:
            while current:
                stack.append(current)
                current = current.left
            current = stack.pop()
            result.append(current.val)
            current = current.right
        return result

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func inorderTraversal(root *TreeNode) []int {
    if root == nil {
        return []int{}
    }
    var result []int
    result = append(result, inorderTraversal(root.Left)...)
    result = append(result, root.Val)
    result = append(result, inorderTraversal(root.Right)...)
    return result
}

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func inorderTraversal(root *TreeNode) []int {
    if root == nil {
        return []int{}
    }

    var result []int
    var stack []*TreeNode
    current := root

    for current !=nil || len(stack) > 0 {
      for current != nil {
        stack = append(stack, current)
        current = current.Left
      }
      current = stack[len(stack) - 1]
      stack = stack[:len(stack) - 1]
      result = append(result, current.Val)
      current = current.Right
    }

    return result
}

94. Binary Tree Inorder Traversal ​

94. Binary Tree Inorder Traversal