如何在js中跳出bts的traverseInOrder方法中的函数？

class TreeNode {
  constructor(value) {
    this.value = value;
    this.left = null;
    this.right = null;
  }
}

function traverseInOrder(node, callback) {
  if (node === null) return;

  traverseIn => Order(node.left, callback);
  callback(node.value);
  traverseInOrder(node.right, callback);
}

// 示例用法
const root = new TreeNode(10);
root.left = new TreeNode(5);
root.right = new TreeNode(15);
root.left.left = new TreeNode(3);
root.left.right = new TreeNode(7);

let shouldBreak = false;

traverseInOrder(root, (value) => {
  console.log(value);

  // 当遇到值为7时，跳出循环
  if (value === 7) {
    shouldBreak = true;
  }

  if (shouldBreak) {
    throw new Error('Break');
  }
});

function traverseInOrder(node, callback) {
  if (node === null) return;

  const stack = [];
  let currentNode = node;

  while (currentNode !== null || stack.length > 0) {
    while (currentNode !== null) {
      stack.push(currentNode);
      currentNode = currentNode.left;
    }

    currentNode = stack.pop();
    if (!callback(currentNode.value)) {
      break;
    }
    currentNode = currentNode.right;
  }
}

// 示例用法
const root = new TreeNode(10);
root.left = new TreeNode(5);
root.right = new TreeNode(15);
root.left.left = new TreeNode(3);
root.left.right = new TreeNode(7);

traverseInOrder(root, (value) => {
  console.log(value);

  // 当遇到值为7时，跳出循环
  return value !== 7;
});